UC-NRI 


V 


- 


UNIVERSITY  EDITION. -REVISED  AND  ENLARGED. 


THEORETICAL  AND  PRACTICAL 


TREATISE    ON 


A    L    G    E    B    R    A: 


IN   WHICH   THE   EXCELLENCIES    OF   THE   DEMONSTRATIVE   METHODS   OF   THE 

FRENCH   ARE   COMBINED   "WITH   THE   MORE   PRACTICAL   OPERATIONS 

OF   THE   ENGLISH  J   AND   CONCISE   SOLUTIONS  POINTED 

OUT    AND   PARTICULARLY   INCULCATED. 


DESIGNED 


SCHOOLS,  COLLEGES,  AND  PRIVATE  STUDENTS. 


BY  H.  N.  ROBINSON,   A.   M. 

FORMERLY  PROFESSOR  OF  MATHEMATICS  IN  THE  UNITED  STATES  NAVY  J   AUTHOR 

OF  A  TREATISE  ON  ARITHMETIC,  ASTRONOMY,  GEOMETRY,  TRIGONOMETRY, 

SURVEYING,  CALCULUS,  NATURAL  PHILOSOPHY,   ctC.  AC. 


TWENTY-EIGHTH   STANDARD    EDITION. 

CINCINNATI: 
PUBLISHED    BY    JACOB    ERNST 

ANDERSON,  GATES  &  WRIGHT,  112  MAIN  STREET. 

D.  ANDERSON   &  CO.:   TOLEDO,  OHIO. 
IVISON  &  PHINNEY,  N.  YORK. 

1858. 


I 
lift  - 


Entered,  according  to  Act  of  Congress,  in  the  year  1847, 

BY  HORATIO  N.ROBINSON, 
In  the  Clerk's  Office  of  the  District  Court  of  the  State  of  Ohio. 


••according  to  aa  of  Congress,  in  the  year  1857, 

BY  H.  N.  ROBINSON, 

in  the  Clerk's  Office  of  the  District  Court  of  the  United  States  far  the  Northern 
District  of  New  York. 


EDUCATION  OEPT 


STEREOTYPED  BY  JAMH8  &  OU., 
CINCINNATI. 


PREFACE. 


SOME  apology  may  appear  requisite  for  offering  a  new  book  to  the  public  on 
the  science  of  Algebra — especially  as  there  are  several  works  of  acknowledged 
merit  on  that  subject  already  before  the  public,  claiming  attention. 

But  the  intrinsic  merits  of  a  book  are  not  alone  sufficient  to  secure  its  adop- 
tion, and  render  it  generally  useful.  In  addition  to  merit,  it  must  be  adapted 
to  the  general  standard  of  scientific  instruction  given  in  our  higher  schools ;  it 
must  conform  in  a  measure  to  the  taste  of  the  nation,  and  correspond  with  the 
general  spirit  of  the  age  in  which  it  is  brought  forth. 

The  elaborate  and  diffusive  style  of  the  French,  as  applied  to  this  science, 
can  never  be  more  than  theoretically  popular  among  the  English;  and  the  se- 
vere, brief,  and  practical  methods  of  the  English  are  almost  intolerable  to  the 
French.  Yet  both  nations  can  boast  of  men  highly  pre-eminent  in  this  science, 
and  the  high  minded  of  both  nations  are  ready  and  willing  to  acknowledge  the 
merits  of  the  other  ;  but  the  style  and  spirit  of  their  respective  productions  are 
necessarily  very  different. 

In  this  country,  our  authors  and  teachers  have  generally  adopted  one  or  the 
other  of  these  schools,  and  thus  have  brought  among  us  difference  of  opin- 
ion, drawn  from  these  different  standards  of  measure  for  true  excellence. 

Very  many  of  the  French  methods  of  treating  algebraic  science  are  not  to  b- ; 
disregarded  or  set  aside.  First  principles,  theories  and  demonstrations,  are  tb  3 
essence  of  all  true  science,  and  the  French  are  very  elaborate  in  these.  Yet  no 
effort  of  individuals,  and  no  influence  of  a  few  institutions  of  learning,  can 
change  the  taste  of  the  American  people,  and  make  them  assimilate  to  the 
French,  any  more  than  they  can  make  the  entire  people  assume  French  ^  iva- 
city,  and  adopt  French  manners. 

Several  works,  modified  from  the  French,  have  had,  and  now  have  consid- 
erable popularity,  but  they  do  not  naturally  suit  American  pupils.  They  are 
not  sufficiently  practical  to  be  unquestionably  popular ;  and  excellent  as  they 
are,  they  fail  to  inspire  that  enthusiastic  spirit,  which  works  of  a  more  practi- 
cal and  English  character  are  known  to  do. 

At  the  other  extreme  are  several  English  books,  almost  wholly  practical,  with 
little  more  than  arbitrary  rules  laid  down.  Such  books  may  in  time  make 
good  resolvers  of  problems,  but  they  certainly  fail  in  most  instances  to  make 
scientific  algebraists. 

The  author  of  this  work  has  had  much  experience  as  a  teacher  of  algebra 

961661 


iv 


PREFACE 


and  has  used  the  different  varieties  of  text  books,  with  a  view  to  test  their  co  n 
parative  excellencies,  and  decide  if  possible  on  the  standard  most  proper  to  be 
adopted,  and  of  course  he  designed  this  work  to  be  such  as  nis  experience  arid 
judgment  would  approve. 

One  of  the  designs  of  this  book  is  to  create  in  the  minds  of  the  pupils  a  love 
for  the  study,  which  must  in  some  way  be  secured  before  success  can  be  at^ 
tained.  Small  works  designed  for  children,  or  those  purposely  adapted  to  per- 
sons of  low  capacity,  will  not  secure  this  end.  Those  who  give  tone  to  public 
opinion  in  schools,  will  look  down  upon,  rather  than  up  to,  works  of  this  kind, 
and  then  the  day  of  their  usefulness  is  past.  On  the  other  hand,  works  of  a 
high  theoretical  character  are  apt  to  discourage  the  pupil  before  his  acquire- 
ments enable  him  to  appreciate  them,  and  on  this  account  alone  such  works 
are  not  the  most  proper  for  elementary  class  books. 

This  work  is  designed,  in  the  strictest  sense,  to  be  both  theoretical  and  prac- 
tical, and  therefore,  if  the  author  has  accomplished  his  design,  it  will  be  fouru\ 
about  midway  between  the  French  and  English  schools. 

In  this  treatise  will  be  found  condensed  and  brief  modes  of  operation,  no» 
nitherto  much  known  or  generally  practised,  and  several  expedients  are  system- 
atised  and  taught,  by  which  many  otherwise  tedious  operations  are  avoided. 

Some  applications  of  the  celebrated  problems  of  the  couriers,  and  also  of  the 
lights,  are  introduced  into  this  work,  as  an  index  to  the  pupil  of  the  subsequent 
utility  of  algebraic  science,  which  may  allure  him  on  to  more  thorough  investi- 
gations, and  more  extensive  study. 

Such  problems  would  be  more  in  place  in  text  books  on  natural  philosophy 
and  astronomy  than  in  an  elementary  algebra,  but  the  almost  entire  absence  of 
them  in  works  of  that  kinu,  is  our  apology  for  inserting  them  here,  if 
apology  be  necessary. 

Quite  young  pupils,  and  such  as  may  not  hare  an  adequate  knowledge  of 
physics  and  the  general  outlines  of  astronomy,  may  omit  these  articles  of  ap- 
plication ;  but  in  all  cases  the  teacher  alone  can  decide  what  to  omit  and  what 
to  teach. 

Within  a  few  years  many  new  text  books  on  algebra  have  appeared  in  differ 
cut  parts  of  the  country,  which  is  a  sure  index  that  something  is  desired — 
something  expected, — not  yet  found.  The  happy  medium  between  the  theo- 
retical and  practical  mathematics,  or,  rather,  the  happy  blending  of  the  two, 
which  all  seem  to  desire,  is  most  difficult  to  attain  ;  hence,  many  have  failed 
in  their  efforts  to  meet  the  wants  of  the  public. 

Metaphysical  theories,  and  speculative  science,  suit  the  meridians  of  France 
and  Germany  better  than  those  of  the  United  States.  But  it  is  almost  impos- 
sible to  comment  on  this  subject  without  being  misapprehended ;  the  author 
of  this  book  is  a  great  admirer  of  the  pure  theories  of  algebraical  science,  for 
it  is  impossible  to  be  practically  skillful  without  having  high  theoretical  acquire- 
ments. It  is  the  man  of  theory  who  brings  forth  practical  respite1  but  it  is  not 
theory  alone — it  is  theory  long  and  wel  applied. 


PREFACE  v 

Who  will  contend  that  Watt,  Fitch,  or  Fulton,  \vcre  ignorant  or  inattentive 
to  every  theory  concerning  the  nature  and  power  of  steam,  yet  they  are  only 
known  as  practical  men,  and  it  is  almost  in  vain  to  look  for  any  benefactors  of 
mankind,  or  any  promoters  of  real  science  from  those  known  only  as  theo- 
rists, or  among  those  who  are  strenuous  contenders  for  technicalities  and  forms. 

We  are  led  to  these  remarks  to  counteract,  in  some  measure,  if  possible,  that 
false  impression  existing  in  some  minds,  that  a  high  standard  work  on  aigebra, 
must  necessarily  be  very  formal  in  manner  and  abstrusely  theoretical  in  mat- 
ter ;  but  in  our  view  these  are  blemishes  rather  than  excellencies. 

The  author  of  this  work  is  a  great  advocate  for  brevity,  when  not  purchased 
at  the  expense  of  perspicuity,  and  this  may  account  for  the  book  appearing 
very  small,  considering  what  it  is  claimed  to  contain.  For  instance,  we  have 
only  two  formulas  in  arithmetical  progression,  and  some  authors  have  20. 
We  contend  the  two  are  sufficient,  and  when  well  understood  cover  the  whole 
theory  pertaining  to  the  subject,  and  in  practice,  whether  for  absolute  use  or 
lasting  improvement  of  the  mind,  are  far  better  than  20.  The  great  number 
onlv  serves  to  confuse  and  distract  the  mind  ;  the  two  essential  ones,  can  be 
remembered  and  most  clearly  and  philosophically  comprehended.  The  same 
remarks  apply  to  geometrical  progression. 

In  the  general  theory  of  equations  of  the  higher  degrees  this  work  is  not  too 
diffuse  ;  at  the  same  time  it  designs  to  be  simple  and  clear,  and  as  much  is 
given  as  in  the  judgment  of  the  author  would  be  acceptable,  in  a  work  as  ele- 
mentary and  condensed  as  this ;  and  if  every  position  is.  not  rigidly  demonstra- 
ted, nothing  is  left  in  obscurity  or  doubt. 

We  have  made  special  effort  to  present  the  beautiful  theorem  of  Sturm  in 
such  a  manner  as  to  bring  it  direct  to  the  comprehension  of  the  student,  and  if 
we  have  failed  in  this,  we  stand  not  alone. 

The  subject  itself,  though  not  essentially  difficult,  is  abstruse  for  a  learner, 
and  in  our  effort  to  render  it  clear  we  have  been  more  circuitous  and  elaborate 
than  we  had  hoped  to  have  been,  or  at  first  intended. 

We  may  apply  the  same  remarks  to  our  treatment  of  Horner's  method  of 
solving  the  higher  equations. 

Brevity  is  a  great  excellence,  but  perspicuity  is  greater,  and,  as  a  goncral 
tiling,  the  two  go  hand  in  hand;  and  these  views  have  guided  us  in  preparing 
the  whole  work ;  we  have  felt  bound  to  be  clear  and  show  the  rationale  of 
every  operation,  and  the  foundation  of  every  principle,  at  whatever  cost. 

The  Indeterminate  and  Diophantine  analysis  are  not  essential  in  a  regular 
course  of  mathematics,  and  it  has  rot  been  customary  to  teach  them  in  many 
institutions ;  for  these  reasons  we  d  >  not  insert  them  in  our  text  book.  The 
teacher  or  the  student,  however,  will  find  the-m  in  a  concise  form  in  a  key  to 
thin  work. 


PREFACE    TO    THE    27TH  (ENLARGED)  EDITION. 


THIS  is  a  progressive  age,  and  teachers,  and  schools,  and  books,  rnusl 
progress,  proportionally,  or  be  left  in  the  rear. 

To  keep  up  with  the  spirit  of  the  times,  and  make  this  volume  more 
complete  and  valuable,  we  have  added  some  twenty-four  pages  of  what  we 
conceive  to  be  very  valuable  matter. 

In  the  demonstration  of  the  binomial  theorem  we  originally  adopted 
that  method,  which,  to  judge  from  our  own  personal  experience  and  obser- 
vation as  a  teacher  of  others,  was  the  one  most  easily  comprehended  by 
learners — but  unfortunately,  that  method  was  not  complete;  it  demonstrated 
only  so  far  as  the  actual  operation  was  carried,  and  the  further  continua- 
tion of  the  law  could  only  be  inferred.  This  being  the  case,  we  have 
added  another  demonstration  in  this  edition,  which  may  require  a  little 
more  mental  discipline  to  clearly  comprehend,  but  which  is,  in  itself,  per- 
fect and  complete. 

We  have  enlarged  the  practical  part  of  logarithms,  and  inserted  two 
small  tables,  which  will  enable  one  to  obtain  the  logarithm  of  any  number 
whatever ;  and  by  the  examples  and  illustrations  here  given,  a  student  can 
obtain  a  very  good  knowledge  of  logarithms,  and  their  uses,  in  a  smaller 
compass  than  can  be  found  in  any  other  book  known  to  us. 

We  have  also  added  several  new  practical  artifices  in  the  solution  of  the 
higher  equations,  which  we  hope  will  not  be  overlooked.  And  to  the 
progressive  teachers  of  the  United  States  the  whole  is  respectfully  dedi- 
cated, by  their  obedient  servant, 

THE  AUTHOR 

EI.CRIDGE,  N".  Y.,  September,  1857. 


VI 


CONTENTS. 

INTRODUCTION 9 

SECTION  I. 

Addition .13 

Subtraction I  (• 

Multiplication 1  !• 

Division 2(> 

Negative  Exponents i27 

ALGEBRAIC  FRACTIONS 34 

Greatest  Common  Divisor '31 

Least  Common  Multiple 42 

Addition  of  Fractions 44 

Subtraction  of  Fractions  46 

Multiplication  of  Fractions 47 

Division  of  Fractions 49 

SECTION  II. 

Equation  of  one  unknown  Quantity 52 

Question  producing  Simple  Equations. 62 

Equations  of  two  unknown  Quantities .70 

Equations  of  three  or  more  unknown  Quantities 76 

Problems  producing  Simple  Equations  of  two  or  more  unknown  quanti- 
ties   87 

Interpretation  of  negative  values  in  the  Solution  of  Problems 94 

Demonstration  of  Theorems 97 

Problem  of  the  Couriers 98 

Application  of  the  Problem  of  the  Couriers 102 

SECTION  III. 

INVOLUTION , 106 

Some  application  of  the  Binomial  Theorem 108 

Evolution 113 

Cube  Root  of  Compound  Quantities 121 

Cube  Root  of  Numerals 123 

Brief  method  of  Approximation  to  the  Cube  Root  of  Numbers 124 

Exponential  Quantities  and  Surds 128 

PURE  EQUATIONS 1 33 

Binomial  Surds 141 

Problem  producing  pure  Equations 147 

Problem  of  the  Lights ..151 

Application  of  the  Problem 152 

vii 


vin  CONTENTS. 

SECTION  IV. 
Quadratic  Equations 157 

Particular  mode  of  completing  a  Square,  (Art.  99) 160 

Special  Artifices  in  resolving  Quadratics,  (Art.  106) 169 

Quadratic  Equations  containing  two  or  more  unknown  Quantities 175 

Questions  producing  Quadratic  Equations 183 

SECTION  V. 

A  rithmetical  Progression ]  89 

Geometrical  Progression 1 95 

flarmonical  Proportion 199 

Problems  in  Progression  and  Harmonical  Proportion 200 

Geometrical  Proportion 205 

SECTION  VI. 

Binomial  Theorem— its  Demonstration,  &c 218 

'•  "  its  General  Application 223 

Infinite  Series 227 

Reversion  of  a  Series 234 

Exponential  Equations  and  Logarithms 239 

Use  and  application  of  Logarithms 256 

Compound  Interest  and  Annuities 258 

SECTION  VII. 

General  Theory  of  Equations 263 

Newton's  Method  of  Division,  (Art.  167) 276 

Equal  Roots 279 

Transformation  of  Equations,  (Art.  171) 282 

Synthetic  Division 290 

Sturm's  Theorem 309 

Newton's  method  of  Approximation 317 

Homer's  Method 31 9 

Its  application  to  Numerical  Roots 333 

Expedients  to  be  used  in  solving  Equations 337 

Recurring  and  Binomial  Equations 340 

Practical  examples  for  solution 343 

General  method  of  Elimination  (extra) 345 

APPENDIX. 

Inequality 349 

Differential  method  of  Series , 350 

Specific  Gravity 357 

Maxima  and  Minima 359 


ELEMENTS  OF  ALGEBRA. 


INTRODUCTION. 


DEFINITIONS    AND    AXIOMS. 

ALGEBRA  is  a  general  kind  of  arithmetic,  an  universal  analysts, 
or  science  of  computation  by  symbols. 

Quantity  or  magnitude  is  a  general  term  applied  to  everything 
which  admits  of  increase,  diminution,  and  measurement. 

The  measurement  of  quantity  is  accomplished  by  means  of  an 
assumed  unit  or  standard  of  measure  ;  and  the  unit  must  be  the 
same,  in  kind,  as  the  quantity  measured.  In  measuring  length,  we 
apply  length,  as  an  inch,  a  yard,  or  a  mile,  &c. ;  measuring  area, 
we  apply  area,  as  a  square  inch,  foot,  or  acre ;  in  measuring 
money,  a  dollar,  pound,  &c.,  may  be  taken  for  the  unit. 

Numbers  represent  the  repetition  of  things,  and  when  no  ap- 
plication is  made,  the  number  is  said  to  be  abstract.  Thus  5,  13, 
200,  &c.,  are  numbers,  but  $5,  13  yards,  200  acres,  are  quanti- 
ties. 

In  algebraical  expressions,  some  quantities  may  be  known, 
others  unknown ;  the  known  quantities  are  represented  by  the 
first  or  leading  letters  of  the  alphabet,  a,  b,  c,  d,  &c,,  and  the 
unknown  quantities  by  the  final  letters,  z,  y,  x,  u,  &c. 

THE    SIGNS. 

(1)  The  perpendicular  cross,  thus  +,  called  plus,  denotes  ad' 
dition,  or  a  positive  value,  state,  or  condition. 

(2)  The  horizontal  dash,  thus  — ,  called  minus,  denotes  siib- 
traction,  or  a  negative  value,  state,  or  condition. 

(3)  The  diamond  cross, thus  X,  or  a  point  between  two  quan- 
tities, denotes  that  they  are  to  be  multiplied  together. 


UO:  ; V !  J  v  INTRODUCTION. 

(4)  A  horizontal  line  with  a  point  above  and  below,  thus  -~, 
denotes  division.     Also,  two  quantities,  one   above  another,  as 
numerator  and  denominator,  thus  ~,  indicates  that  a  is  divided 
by  b. 

(5)  Double  horizontal  lines,  thus  =,  represent  equality.  Points 
between  terms,  thus  a  :  b  ::  c  :  d,  represent  proportion,  and  ar« 
read  as  a  is  to  b  so  is  c  to  d. 

(6)  The  following  sign  represents  root  J ;  alone  it  signifies 
square  root.     With  small  figures  attached,  thus  *.J  4,J  5N/,  &c., 
indicates  the  third,  fourth,  fifth,  &-.,  root. 

Roots  may  also  be  represented  by  fractions  written   over  a 

quantity,  as  aa  «3  «?,  &c.,  which   indicate   the  square   root, 
the  third  root,  and  fourth  root  of  «.* 

(7)  This  symbol,  of>6,  signifies  that  a  is  greater  than  b. 
This       «*      ,  a<^b,  signifies  that  a  is  less  than  b. 

(8)  A  vinculum  or  bar ,  or  parenthesis  (  )  is  used  to  con- 
nect several  quantities  together.  Thus  «-j-6-{-cX#,or  (a-}-b-{-c)x, 
denotes  that  a  plus  b  plus  c  is  to  be  multiplied  by  x.     The  bar 

may  be  placed  vertically,  thus,  which  is   the   same 


as  (a — d-\-e}y,  or  the  same  as  ay — dy-\-ty  without  the 


— d 

+4 

vinculum. 

(9)  Simple  quantities  consist  of  a  single  term,  as  a,  b,  ab,  3#, 
&c.     Compound  quantities  consist  of  two  or  more  terms  coli- 
nected  by  their  proper  signs,  as  a-\-x,  3b-\-2y,  7ab — 3xy-{-c,  <fcc 
A  binomial  consists  of  two  terms  ;  a  trinomial  of  three ;  and  a 
polynomial  of  many,  or  any  number  of  terms  above  two. 

(10)  Quantities  multiplied  together  are  called  factors  of  the 
product,  and  factors  are  called  coefficient  in  respect  to  each  other. 
Thus,  in  the  expressions  3#,  abx ;  3  and  x  are  factors  of  the 
quantity  3a?,  and  3  is  the  coefficient  of  x.     Also,  a,  b,  and  x  are 
factors  of  abx ;   a  is  the  coefficient  of  bx ;  b  is  the  coefficient 
of  ax  ;  and  ab  is  the  coefficient  of  x. 

(11)  The  measure  of  a  quantity  is  some  exact  fact  or  of  that 
quantity :  thus,  5«  is  the  measure  of  20«  or  25«. 

*The  adoption  and  utility  of  this  last  mode  of  notation,  which  ought  to 
be  exclusively  used,  will  be  explained  in  a  subsequent  part  of  this  worU. 


EXERCISES  ON  NOTATION.  u 

(12)  The  root  of  any  quantity  is  some  equal  factor  of  that 
quantity.     The  square  root  is  one  of  two  equal  factors  ;  the  cube 
root  one  of  three  equal  factors,  and  so  on. 

(13)  The  multiple  of  any  quantity  is  some  exact  number  of 
limes  that  quantity. 

Other  terms  will  be  defined  as  we  use  them. 
AXIOMS. 

Axioms  are  self-evident  truths,  and  of  course  are  above  demon 
stration  ;  no  explanation  can  render  them  more  clear.     The  fol- 
lowing are  those  applicable  to  algebra,  and  are  the  principles  on 
which  the  truth  of  all  algebraical  operations  finally  rests: 

Axiom  1.  If  the  same  quantity  or  equal  quantities  be  added  to 
equal  quantities,  their  sums  will  be  equal. 

2.  If  the  same  quantity  or  equal  quantities  be  subtracted  from 
equal  quantities,  the  remainders  will  be  equal. 

3.  If  equal  quantities  be  multiplied  into  the  same,  or  equal 
quantities,  the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same,  or  by  equal 
quantities,  the  quotients  will  be  equal. 

5.  If  the  same  quantity  be  both  added  to  and  subtracted  from 
another,  the  value  of  the  latter  will  not  be  altered. 

6.  If  a  quantity  be  both  multiplied  and  divided  by  another, 
the  value  of  the  former  will  not  be  altered. 

7.  Quantities  which  are  respectively  equal  to  any  other  quan- 
tity are  equal  to  each  other. 

8.  Like  roots  of  equal  quantities  are  equal. 

9.  Like  powers  of  the  same  or  equal  quantities  are  equal. 


EXERCISES  ON  NOTATION. 

When  definite  values  are  given  to  the  letters  employed,  we 
can  at  once  determine  the  value  of  their  combination  in  any  alge- 
braic expression 


12  ELEMENTS  OF  ALGEBRA. 

Let  a=5     6=20     c=4     d=l 
Then  a+6—  •  c=5+20—  4     or  a+6—  c=21 


a&+aH-d=5X  20+5X4+1  =  121 


SECTION    I. 

ADDITION. 

(Art.  1.)  Before  we  can  make  use  of  literal  or  algebraical 
quantities  to  aid  us  in  any  mathematical  investigation,  we  must 
not  only  learn  the  nature  of  the  quantities  expressed,  but  how  to 
add,  subtract,  multiply,  and  divide  them,  and  subsequently  learn 
how  to  raise  them  to  powers,  and  extract  roots. 

The  pupil  has  undoubtedly  learned  in  arithmetic,  that  quanti- 
ties representing  different  things  cannot  be  added  together ;  for 
instance,  dollars  and  yards  of  cloth  cannot  be  put  into  one  sum ; 
but  dollars  can  be  added  to  dollars,  and  yards  to  yards  ;  units  can 
be  added  to  units,  tens  to  tens,  &c.  So  in  algebra,  a  can  be 
added  to  a,  making  2a ;  3a  can  be  added  to  5a,  making  Sa.  As 
a  may  represent  a  dollar,  then  3a  would  be  3  dollars,  and  5« 
would  be  5  dollars,  and  the  sum  would  be  8  dollars.  Again,  a 
may  represent  any  number  of  dollars  as  well  as  one  dollar ;  for 
example,  suppose  a  to  represent  6  dollars,  then  3a  would  be  18 
dollars,  and  5a  would  be  30  dollars,  and  the  whole  sum  would  be 
48  dollars.  Also,  8«  is  8  times  6  or  48  dollars  ;  hence  any  num- 
ber of  «'s  may  be  added  to  any  other  number  of  «'s  by  uniting 
their  coefficients ;  but  a  cannot  be  added  to  &,  or  4«  to  36,  or  to 
any  other  dissimilar  quantity,  because  it  would  be  adding  unlike 
things,  but  we  can  write  a-\-b  and  3«+36,  indicating  the  addi- 
tion by  the  sign,  making  a  compound  quantity, 


ADDITION.  13 

Let  the  pupil  observe  that  a  broad  generality,  a  wide  latitude 
must  be  given  to  the  term  addition.  In  algebra,  it  rather  means 
uniting-,  condensing,  or  reducing  terms,  and  in  some  cases,  the 
sum  may  appear  like  difference,  owing  to  the  difference  of  signs. 
Thus,  4a  added  to  — a  is  3a ;  that  is,  the  quantities  wilted  can 
make  only  3a,  because  the  minus  sign  indicates  that  one  a  must 
be  taken  out.  Again,  76-f-36 — 46,  when  united,  can  give  only 
66,  which  is  in  fact  the  sum  of  these  quantities,  as  4b  has  the 
minus  sign,  which  demands  that  it  should  be  taken  out;  hence  to 
add  similar  quantities  we  have  the  following 

RULE.  Add  the.  affirmative,  coefficients  into  one  sum  and  the 
negative  ones  into  another,  and  take  their  difference  with  the 
sign  of  the  greater,  to  which  affix  the  common  literal  quan- 
tity. 

EXAMPLES    FOR    PRACTICE. 

5a         llx         +  5#6  6a-f56         — 7cd  +  8xy 

2a  2x         — Gab         —Ga—4b         — 2cd  +  3xy 

Sum  la         IQx          —ab  +6         —9cd+llxy 


5fl-f& 

cdy+ax 

4x  —  6 

3a-l-c 

2cdy  —  '3  ax 

2^+10 

7o—  26+c 

4cdy-}-3ax 

—  -3x+  7 

—  3a--36—  4c 

—  Icdy  —  ax 

6rc—  12 

Sum        12«—  46—  2c  0  0  9x—  1 

N.  B.  Like  quantities,  of  whatever  kind,  whether  of  powers  or 
roots,  may  be  added  together  the  same  as  more  simple  or  rational 
quantities. 

Thus  3a2  and  8a*  are  11«2,  and  763-f-363=1063.  No  matter 
what  the  terms  may  he,  if  they  are  only  alike  in  kind.  Let  the 
reader  observe  that  2(a-\-b~)-\-3(a-{-b)  must  be  together  5(a-\-b), 
mat  is,  2  times  any  quantity  whatever  added  to  3  times  the 
same  quantity  must  be  5  times  that  quantity.  Therefore, 


{-y=ltJxJry1  for  Jx+y,  which  represents  the 
square  root  of  x-\-y,  may  be  considered  a  single  quantity. 

(Art.  2.)  To  find  the  sum  of  various  quantities  we  have  the 
following 


14  ELEMENTS  OF  ALGEBRA. 

RULE.  Collect  together  all  those  that  are  alike,  by  willing 
their  coefficients,  and  then  write  the  different  sums,  one  after 
another,  with  their  proper  signs. 


EXAMPLES. 


1. 

2. 

3. 

3xy 

,         9x*y 

6xy—  12  x? 

2ax 

—7x*y 

—4^  +3xy 

—5xy 

-\-3axy 

~{-4x*  —  2xy 

Gax 

~4x*y 

-  -3^-{-4a?2 

Sum  Sax  —  2xy 

3axy  —  2x?y 

4xy  —  8^ 

4. 

5. 

6. 

1 4ax — 2  xz  9-j-lO  Jax — 5; y 

5ax-{-3xy  2x-}-7lJxy-{-5y 

Syz — 4ax  5y-\-3,Jax-\-4y 

^4-40—6^               3a^+26  10  —  4Jax+4y 


7.  Add  2xy—2az,   3az-{-xy,  az-\-xy,   4a? — 3xy,    2xy — 2a*. 

Ans.     4a?-{-3xy. 

8.  Add  8aV — 3xy,      Sax — 5xy,     Qxy — 5ax,      2azxz-\-xy, 
5ax — 3xy.  rfns.     I0azxz-+-5ax — xy. 

O.  Add  2a?*— 10y,    3jxy-{-Wx,   2xzy-\-25y,    \2xy—,Jxy, 
—Sy-f-nV^. 

Ans.     2xzy+12xy-{-Wx-\- 

10.  Add  26a>-12,      3x*—2bx,      5a^— 3  Jx, 
a!*+3.  dns. 

11.  Add   1062— 3bx*,      2bzxz—bz,      10—26^,      bzxz— 20, 
3bxz+bz.  Ans.     10^2+362a>— 2bxz— 10. 

12.  Add  2a2— Zax^+x2,     2ax*— -\3xy-\-9,     Wtf—xy — i. 

JJns.     12a2— ax*+x*—  Uxy+4. 

1».  Add  9&c8—18ac*      156c3--ac     9ac2— 246c3,     9ac2— 2 

.    ac — 2. 


ADDITION.  15 

14.  Add    3m2— 1,     6am  —  2w2-f-4,     7  — 8a?/i-}-2w2,  and 
6ro2-f2am-H.  ^ns      9ro2+ll. 

15.  Add    12a— 13a6-i-16a;r,     8— 4m+2y,    — 6a-f-7a624- 
\%y — 24,  and  lab — lQax-{-4m. 

rfns.     6a— 6a6+ 1 4y -f-7o62—  1 6. 

16.  Add  72ax4— 8a#3,       —  38«z4— 3ay4+7ay*,    S+l2ay\ 
-607/4-12— 34az4+5«i/3— 9oy4.  ^ns.     — 2ai/8+20. 

Add  a-(-6  and  3« — 56  together. 

Add  Go; — 56-f  a+8  to  — 5a—4x-\-4b—3. 

Add  a+26—  3c — 10  to  36 — 4a+5c-}-10  and  56— c. 

Add  3a-f-6 — 10  to  c — d—a  and  — 4c-J-2« — 36 — 7. 

Add  3a2-l-262— c  to  206— 3«2-{-6c— 6. 

(Art.  3.)  When  similar  quantities  have  literal  coefficients,  we 
may  add  them  by  putting  their  coefficients  in  a  vinculum,  and 
writing  the  term  on  the  outside  as  a  factor. 

Thus  the  sum  of  ax  and  bx    is 


Add 

EXAMPLES. 
1. 

ax+bif 

2. 

ay+cx 
3ay-{-2cx 
4y  -{-Qx 

Sum  («-}-- 
Add 

lc+4d)x-\-(b-\-3a-{-7)y2        ( 

3. 

3x-\-2xy 
bx-^cxy 
(a  -\-b}x-{-2cdxy 

4. 

ax-}-7y 
7ax  —  3y 
—  2x  -\-4y 

Sum 

5.  Add  8«oH-2(;r+a)-r-36,    9ax+6(x-{-a)—  96,    and 
66—  7ax—  8(a?+«). 


6.  Add  (a-\-b),Jx  and  (c+2a  —  6)7a?  together. 

7.  Add  28o3(aM-57/)+21,     18«—  13a3(#-f-5i/), 

.     18a+13. 


16  ELEMENTS  OF  ALGEBRA. 

8.  Add  17a(x-h3«t/)-J-12aW,  8— 18ay— 


SUBTRACTION. 

(Art.  4.)  We  do  not  approve  of  the  use  of  the  term  subtraction, 
as  applied  to  algebra,  for  in  many  cases  subtraction  appears  like 
addition,  and  addition  like  subtraction.  We  prefer  to  use  the 
expression^/u/mg1  the  difference. 

What  is  the  difference  between  12  and  20  degrees  of  north 
latitude  ?  This  is  subtraction.  But  when  we  demand  the  differ- 
ence of  latitude  between  6  degrees  north  and  3  degrees  south,  the 
result  appears  like  addition,  for  the  difference  is  really  9  de- 
grees, the  sum  of  6  and  3.  This  example  serves  to  explain  the 
true  nature  of  the  sign  minus.  It  is  merely  an  opposition  to 
the  sign  plus ;  it  is  counting  in  another  direction;  and  if  we 
call  the  degrees  north  of  the  equator  plus,  we  must  call  those 
south  of  it  minus,  taking  the  equator  as  the  zero  line. 

So  it  is  on  the  thermometer  scale ;  the  divisions  above  zero 
are  called  p!ust  those  below  minus.  Momsy  due  to  us  may  be 
called  plus  ;  money  that  we  owe  should  then  be  called  minus, — 
the  one  circumstance  is  directly  opposite,  in  effect,  to  the  other. 
Indeed,  we  can  conceive  of  no  quantity  less  than  nothing,  as 
we  sometimes  express  ourselves*  It  is  quantity  in  opposite  cir- 
cumstances or  counted  in  an  opposite  direction;  hence  the  differ' 
ence  or  space  between  a  positive  and  a  negative  quantity  is 
their  apparent  sum. 

As  a  further  illustration  of  finding  differences,  let  us  take  the 
following  examples,  which  all  can  understand : 

From       16  16  16  16  16  16 

Take       12  8  2  0  —2  —4 

Differ,       4  8  14  16  18  ~         20 

Here  the  reader  should  strictly  observe  that  the  smaller  the 
number  we  take  away,  the  greater  the  remainder,  and  when  the 
subtrahend  becomes  minus,  it  must  be  added. 


SUBTRACTION.  17 

From        I2a  120          12a          120          120          12a 

Take        20«  16«          12a  9«  60  a 

Diff.       _8«        —  4a  0  30  60         lla 

When  a  greater  is  taken  from  a  less,  we  cannot  have  &  posi- 
tive or  plus  difference,  it  must  be  minus. 

From  200  100  50  0  0  —  50  —  100 
Take  lla  lla  lla  lla  —  b  —  b  —50 
Diff.  90  —  a  —60  —lla  +b  b—5a  —50 

Here  it  will  be  perceived,  that  any  quantity  subtracted  from 
zero  will  be  the  same  quantity  with  its  sign  changed. 

(Art.  5.)  Unlike  quantities  cannot  be  written  in  one  sum,  (Art 
1,)  but  must  be  taken  one  after  another  with  their  proper  signs  : 
therefore,  the  difference  of  unlike  quantities  can  only  be  ex- 
pressed by  signs.  Thus  the  difference  between  a  and  b  is  a  —  ft, 
a  positive  quantity  if  a  is  greater  than  6,  otherwise  it  is  negative 
From  a  take  b  —  c,  (observe  that  they  are  unlike  quantities). 

OPERATION. 
From  a-fO-f-0 

Take  Q+b—  c 

Remainder,  or  difference,  a  —  6-f-c 

This  formal  manner  of  operation  may  be  dispensed  with  ;  the 
ciphers  need  not  be  written,  and  the  signs  of  the  subtrahend  need 
only  be  changed. 

From  the  preceding  observation,  we  draw  the  following 

GENERAL  RULE  FOR  SUBTRACTION,  OR  ALGEBRAIC  DIF- 

FERENCES. 

Change  the  signs  of  the  subtrahend,  or  conceive  them  to  be 
clanged;  then  proceed  as  in  addition. 

EXAMPLES. 

1.                                2.  3. 

From              4a+2x—  3c                3ax-{-2y  a-\-b 

Take                 a-\-4x  —  6c                  xy—ly  a  —  & 


Remainder,      3a  —  2a?+3c  Sax  —  xy-\-^y  26 


18  ELEMENTS  OF  ALGEBRA. 

4.                                  5.  6. 

From        2#2 — 3x-^-y2             7a+2 — 5c  < 

Take       — x* — 4aH-a             — a-J-2-1-  c  %x — 


Rem.      3r>-l-#--2— a  8a   *   — 6c 


7.  8. 

Fmm  8x2 — 3xy-}-2yz-}-  c  ax-\-bx-\-cx 

Take  a,'2 — Qxy-\-3yz — 2c 

Diff. 


9. 

From  ax-\-by-{-cz 

Take  —mx — ny — pz 


Diff. 


(Art.  6.)  From  a  take  b.  The  result  is  « — 6.  The  minus 
sign  here  shows  that  the  operation  has  been  performed ;  b  was 
positive  before  the  subtraction ;  changing  the  sign  performed 
the  subtraction;  so  changing  the  sign  of  any  other  quantity 
would  subtract  it. 

11.  From  3«  take  (ab-\-x — c — ^/),  considering  the  terms  in 
the  vinculum  as  one  term,  the  difference  must  be  3a — (ab-\-x — 
c — y),  but  if  we  subtract  this  quantity  not  as  a  whole,  but  term 
by  term,  the  remainder  must  be  3a — ab — aj-j-c-f-y. 

That  is,  when  the  vinculum  is  taken  away,  all  the  signs 
within  the  vinculum  must  be  changed. 

12.  From  3Qxy  take  (40#y— 262+3c— 4d). 

Rem.     2b2 — IQxy — 3c-{-4c/. 


13.  From  Jx-\-y+3ax— 12  take 

Rem.     5ax — 3jx-}-v— 12— b. 


14.  Find  the  difference  between  6?/2 — 2i/ — 5   and  — 8^/2 — 
4-12. 

15.  From  3a— 6— 2o;+7  take 


MULTIPLICATION.  19 

10.  From  3p-\-q-\-r  —  25  take  q—8r-}-2s  —  8. 

dns.     3j»-f-9r—  4s+8 

17.  From  13a2—  2ax+9x*  take  5az—  lax—  x2. 

An*.     Sa*+5ar-\-lOx* 
\_ 

18.  From  20xy  —  S^a-f  3i/  take  to/+5a2—  y 


19.  From  the  sum  of    Gorfy—  Ilax*,   and  8;r2i/+3a;r3,  take 
42^  —  4ax^ttf  Diff. 

20.  From  the  sum  of  15a26-r-8ctfo  —  3  and 

take  the  sum  of  12a2&  —  3cdx  —  8  and  16+cdx  —  4azb. 

Diff. 


21.  From  the  difference  between   Sab  —  I2cy  and  —  3ab-\- 
y  take  tlie  sum  of  Sab  —  Icy  and  ab-\-cy» 

Diff.     5ab  —  Wcy. 
From  2«+26  take  —  a—b. 
From  ax-}-bx  take  «.T  —  bx. 
From  a-{-c+6  take  a-{-c  —  b. 
From  3a?+2y+2  take  5x-\-3y+b. 
From  6«-f  2j?+c  take  5a+6a?  —  3c. 
From  —  4a  —  2x  —  2  take  —  6a  —  2a?  —  2. 
From  12^—2^7/4-3  take  7+Gy+Wx. 


MULTIPLICATION. 

(A.rt.  7.)  The  nature  of  multiplication  is  the  same  in  arithmetic 
and  algebra.  It  is  repeating  one  quantity  as  many  times  as  there 
are  units  in  another ;  the  two  quantities  may  be  called  factors, 
and  in  abstract  quantities,  either  may  be  called  the  multiplicand ; 
the  other  will  of  course  be  the  multiplier. 

Thus  4X5.  It  is  indifferent  whether  we  consider  4  repeated 
5  times  or  5  repeated  4  times  ;  that  is,  it  is  indifferent  which  we 
call  the  multiplier.  Let  a  represent  4,  and  b  represent  5,  then 
the  product  is  aXb  ;  or  with  letters  we  may  omit  the  sign  and 
the  product  will  be  simply  ab. 


20  ELEMENTS  OF  ALGEBRA. 

The  product  of  any  number  of  letters,  as  ab  c  d,  is  abed. 

The  product  of  x  y  z  is  xyz. 

Tn  the  product  it  is  no  matter  in  what  order  the  letters  are 
placed ;  xy  and  yx  is  the  same  product. 

The  product  of  axXby  is  axby  or  abxy.  Now  suppose 
«=6  and  6=8,  then  ab=4S,  and  the  product  of  axXby  would 
be  the  same  as  the  product  of  GxXSy  or  4Sxy.  From  this  we 
draw  the  following  rule  for  multiplying  simple  quantities : 

Multiply  the  coefficients  together  and  annex  the  letters,  one 
after  another,  to  the  product. 

EXAMPLES. 

1.  Multiply  3x  by  7 a.  Prod.     21  ax. 

2.  Multiply  4y  by  Sab  Prod.     IZaby. 

3.  Multiply  36  by  5c,  and  that  product  by  Wx. 

Prod.     ISQbcx 

4.  Multiply  Qax  by  I2by  by  7ad.         Prod.     5Q4aaxydb. 

5.  Multiply  Sac  by  116  by  xy. 

6.  Multiply  af  by  pq  by  4. 

In  the  above  examples  no  signs  were  expressed,  and  of  course 
plus  was  understood ;  and  it  is  as  clear  as  an  axiom  that  plus 
multiplied  by  plus  must  produce  plus,  or  a  positive  product. 

(Art.  8.)  As  algebraic  quantities  are  liable  to  be  affected  by 
negative  signs,  we  must  investigate  the  products  arising  from 
them.  Let  it  be  required  to  multiply  — 4  by  3,  that  is,  repeat 
the  negative  quantity  3  times ;  the  whole  must  be  negative,  as 
the  sum  of  any  number  of  negative  quantities  is  negative. 
Hence  minus  multiplied  by  plus  gives  minus,  — aXb  gives 
— ab  ;  also  a  multiplied  by  — b  must  give  — ab,  as  we  may 
Conceive  the  minus  b  repeated  a  times. 

(Art.  9.)  Now  let  it  be  required  to  multiply  — 4  by  — 3,  that 
is,  minus  4  must  be  subtracted  3  times ;  but  to  subtract  minus 
4  is  the  same  as  to  add  4,  (Art.  5,)  giving  a  positive  or  plus 
quantity;  and  to  subtract  it  3  times,  as  the  — 3  indicates,  will 
give  a  product  of  +12. 

That  is,  minus  multiplied  by  minus  gives  phis. 


MULTIPLICATION.  21 

This  principle  is  so  important  that  we  give  another  mode  of 
illustrating  it: 

Required  the  product  of  a — b  by  a — c. 

Here  a — b  must  be  repeated  a — c  times. 

If  we  take  a — b,  a  times,  we  shall  have  too  large  a  product, 
as  the  multiplier  a  is  to  be  diminished  by  c. 

That  is  a — b 

Multiplied  by      a 

Gives  ua — ab1  which  is  too  great  by  a — b  repeated 

c  times,  or  by  ac — cb,  which  must  be  subtracted  from  the  former 
product;  but  to  subtract  we  change  signs,  (Art.  5,)  therefore  the 
true  product  must  be  aa — ab — ac-}-cb. 

That  is,  the  product  of  minus  b  by  minus  c  gives  plus  be, 
and,  in  general,  minus  multiplied  by  minus  gives  plus. 

But  plus  quantities  multiplied  by  plus  give  plus,  and  minus  by 
plus,  or  plus  by  minus,  give  minus  ;  therefore  we  may  say,  in 
short, 

77m/  quantities  affected  by  like  signs,  when  multiplied  toge- 
ther, give  plus,  and  when  affected  by  unlike  signs  give  minus. 

(Art.  10.)  The  product  of  a  into  b  can  only  be  expressed  by 
ab  or  ba.  The  product  of  abed,  &c.,  is  abed;  but  if  b  c  and  d 
are  each  equal  to  a,  the  product  would  be  aaaa. 

The  product  of  aa  into  aaa  is  aaaaa ;  but  for  the  sake  of 
brevity  and  convenience,  in  place  of  writing  aaa,  we  write  a3. 
The  figure  on  the  right  of  the  letter  shows  how  many  times  the 
letter  is  taken  as  a  factor,  and  is  called  an  exponent.  The  pro- 
duct of  a3  into  a4  is  a  repeated  3  times  as  a  factor,  and  4  times 
as  a  factor,  in  all  7  times  ;  that  is,  write  the  letter  and  add  the 
exponents. 

EXAMPLES. 

What  is  the  product  of  a9  by  a5  ?  Jlns.     a*. 

What  is  the  product  of  x*  by  #8  ?  Am.     xl\ 

What  is  the  product  of  i/2  by  y*  by  t/5?  JJns.    yw. 

What  is  the  product  of  an  by  am  ?  Ans.     a"  + "'. 

>Vhat  is  the  product  of  6V  by  bx.  Ans.     b*x*. 

What  is  the  product  of  ac  by  ac2  by  aV  ?  Ans.    ^5c6. 


22  ELEMENTS  OF  ALGEBRA. 

If  adding  numeral  exponents  is  a  true  operation,  it  must  be 
equally  true  when  the  exponents  are  literal. 

N.  B.  When  the  exponent  is  not  expressed,  one  is  understood, 
for  a  is  certainly  the  same  as  a1,  or  once  taken. 

(Art.  11.)  Every  factor  must  appear  or  be  contained  in  a  pro- 
duct. Thus  axz  multiplied  by  bx3  must  be  abx5.  Now  if  a—  6 
and  b  ==10  the  product  would  be  60#5. 

Multiply  3a2  by  7«3.         Product  21a5. 

From  this  we  draw  the  following  rule  for  the  multiplication 
of  exponential  quantities. 

Multiply  the,  coefficients  and  add  the  exponents  of  the  same 
latter.  Jill  the  letters  must  appear  in  the  product. 

EXAMPLES. 

Multiply  4a4  by  3«.  Prod.     12«5. 

Multiply  3xz  by  — 2x*.  Prod.     — O.r5. 

Multiply  3x  by  7x*  by  3a?y.  Prod.     63aVy. 

What  is  the  product  of  2axz,  4axy,  7abx  ? 

Prod.     5Ga?x*by. 
What  is  the  product  of  *2&n,  3amx,  and  ax  1 

Prod.     Gan+m+\x*. 

Multiply  9a?x  by  4x.  Prod.     36a2.r2. 

Multiply  ITtfW  by  7ac.  Prodf.     119«462c4. 

Multiply  Ilfl5i2c  by  10«5i8c9,  Prod     110a10610c10. 

Multiply  121i2c3^  by  5aAbxy2.  Prod.     605a4i3c3^y. 

Multiply  77a3cz4  by  61«2^.  Prod.     4Q97a5bcx*. 

Multiply  117a62c3a:  by  2«3i2c.  Prod.     234aWc*x. 

Multiply  9azx  by  Gx. 
Multiply  9 ax2  by  — 7 ax. 
Multiply  7 ax  by  — 4. 
Multiply  Sac  by  —  2cx  by  — 4c.  Prod     24«cs.z 

(Art.  12.)  When  one  compound  quantity  is  to  be  multiplied 
or  repeated  as  many  times  as  there  are  units  in  another,  it  is 
evident  that  the  multiplicand  must  be  repeated  by  every  term  of 
the  multiplier. 


MULTIPLICATION. 


Thus  the  product  of  a-\-b-}-c  by  x-\-y+z. 
It  is  evident  that  a-\-b-{-c  must  be  repeated  x  times,  then  y 
times,  then  z  times  ;  and  the  operation  may  stand  thus  : 


Product  by  x    ax-\-bx-\-cx 
Product  by  y  ay-{-by-\-cy 

Product  by  z  az-}-bz-}-cz 


Entire  Product  ax-{-bx-}-cx-{-ay-}-by-\-cy-}-az-{-bz-{-cz. 

From  the  foregoing  articles  we  draw  the  following  general 
rule  for  the  multiplication  of  compound  quantities. 

Multiply  all  the  terms  of  the  multiplicand  by  each  terra  of 
the  multiplier,  observing  that  like  signs,  in  both  factors,  give 
plus,  and  unlike,  minus. 

Write  each  term  of  the  product  distinctly  by  itself,  with  its 
proper  sign,  and  afterwards  condense  or  connect  the  terms  as 
much  as  possible,  as  in  addition. 

EXAMPLES. 

1.  2. 

Multiply  2ax— 3x  3x-\-  2y 

By  2x  -\-4y  4x —  5y 

Partial  product  4ax2-—  6x*  l2x*-\-  Sxy 

2d  part.  prod.  Saxy — I2xy  — 15xy — ICty2 

Whole  prod.      4ax-{-Saxy—6xz—  I2xy     12^—  7xy—Wy* 

3.  Multiply  2xz-{-xy — 2?/2 

By  3x  — 3^ 


Partial  product        Qx*-\-3x~y  —  Gxy2 

2d  partial  product        —  6x*y  —  3xy2-}-6y3 

Whole  product       Gx3  —  3x2y—  9xy*+6y* 


4.  Multiply  3a2—  2a5—  bz  by  2«—  4b. 

'Prod      6a3—  16a26-f-6«&2-f  4#>. 

5.  Multiply  x*—xy-\-if  by  x+y.  Prod, 


24  ELEMENTS  OF  ALGEBRA. 

6.  Multiply  a2— 3ac-f-c2  by  a — c. 

Prod,     a3— 4a2c+4«c2— c8. 

7.  Multiply  a-f-6  by  a-\-b.  Prod.     a2+2a&-f-&2- 

8.  Multiply  x+y  by  #+#•  /Vod.     rc2+2a<i/+3/2. 

9.  Multiply  a — b  by  a— b.  Prod.     a2—2ab-\-b': 

10.  Multiply  x — y  by  x — y.  Prod,     x* — 2xy-\-y\ 

(Art.  13.)  By  inspecting  all  the  problems,  from  the  7th  to  the 
10th,  we  shall  perceive  that  they  are  all  binomial  quantities,  and 
the  multiplicand  and  multiplier  the  same. 

But  when  a  number  is  to  be  multiplied  into  itself  the  product 
is  called  a  square.  Now  by  inspecting  the  products,  we  find 
that  the  square  of  any  binomial  quantity  is  equal  to  plus  ;  the 
squares  of  the  two  parts  and  twice  the  product  of  the  two  parts. 

N.  B.  The  product  of  the  two  parts  will  be  plus  or  minus, 
according  to  the  sign  between  the  terms  of  the  binomial. 

Let  us  now  examine  the  product  of  o-{-b  into  a — b. 
a  -}-b  2m  +2n 

a  — b  2m  — 2n 


a?-\-ab  4m2-\-4mn 

— ab — b2  — 4mn — 4?i2 


Product    a2        — b*  4m*  — in2 

Multiply  20-J-3&  by  2a— 3b.  Prod.     4a*—Qb2. 

Multiply  3y-\-c  by  3y—c.  Prod.     9j/2— c2. 

Thus,  by  inspection,  we  find  the  product  of  the  sum  and 
difference  of  two  quantities  is  tqual  to  the  difference  of  their 
squares. 

The  propositions  included  in  this  article  are  proved  also  in 
geometry. 

(Art.  14.)  We  can  sometimes  make  use  of  binomial  quantities 
greatly  to  our  advantage,  as  a  few  of  the  following  examples  will 
show : 

1.  Multiply  a-r-6-f-c,  by  a+6  f  c. 

Suppose  a-\-b       represented  by  s,  then  it  will  be  s-J-c. 


MULTIPLICATION.  25 

The  square  of  this  is  s2-f-2se-fc- ;  restoring  the  value  of  s,  and 
we  have  (a-}-b)~+2(a+b)c-\-c2. 

2.  Square  x-\-y — z.     Let  x-}-y=s. 

Then  («— z)2=s2 — 2sz+z*=(x-\-y}2— 2(x+y)z-{-z*. 

&.  Multiply  x-\-y-\-z  by  x+y — z.         Prod.    (x-\-y}2—z*. 

4.  Multiplj   2a;2— 3a:+2  by  a?— 8. 

I     2X3— 19^+260;— 10. 


5.  Multiply  ax-\-by  by  ax+cy. 

Prod.     d*x2+(ab-\-ac)xy+cby*. 


6.  Multiply  %x+y  by 

7.  Multiply  a3-f  2a26+2a62  +6 
By  a3— 


Prod. 


8.  Multiply    ^—  ^  a?  +f 
By  \x  +  2 


Product,    la8-}-  V  a^—Js-K 
9.  What  is  the  product  of  am+6m  by  on+6n  ? 


10.  What  is  the  product  of  z2—  fa?  by  a;2—  ia?  ? 

^fns.     as*—  f  a?»+  1  a;2 

11.  What  is  the  product  of  4z3-|-8;r2-r-16aM-32  by  3x—  0? 

4—  192. 


12.  What  is  the  product  of  a*-}-a2b-{-ab2-{-b2  by  a  —  b  ? 

w?ns.     a4—  ** 
3 


26  ELEMENTS  OF  ALGEBRA. 

DIVISION. 

(Art.  15.)  Division  is  the  converse  of  multiplication,  the  pro* 
duct  being  called  a  dividend,  and  one  of  the  factors  a  divisor.  If 
a  multiplied  by  b  give  the  product  «&,  then  ab  divided  by  a  must 
give  b  for  a  quotient,  and  if  divided  by  b,  give  a.  In  short,  if 
one  simple  quantity  is  to  be  divided  by  another  simple  quantity, 
the  quotient  must  be  found  by  inspection,  as  in  division  of  num- 
bers. 

EXAMPLES. 

1.  Divide  IGab  by  4a.  *An&.     4b 

2.  Divide  2  lacd  by  7c.  rfns.     3a(l 

3.  Divide  ab*c  by  ac.  Am.     62. 

4.  Divide  Zaxy  by  2bc.  Ans.     ?^ 

In  this  last  example,  and  in  many  others,  the  absolute  division 
cannot  be  effected.  In  some  cases  it  can  be  partially  effected, 
and  the  quotients  must  be  fractional. 

5.  Divide  3acx*  by  ac?/.  Ans.     -  . 

*  J  y 

Qb 

6.  Divide  72te  by  8abx.  Ans.     —  . 


7.  Divide  27aby  by  llabx.  Ans. 

(Art.  16.)  It  will  be  observed  that  the  product  of  the  divisoi 
and  quotient  must  make  the  dividend,  and  the  signs  must  con- 
form to  the  principles  laid  down  in  multiplication.  The  follow- 
ing examples  will  illustrate  : 

8.  Divide  —  9y  by  3?/.  jlns.    —  3. 

9.  Divide  —  9i/  by  —  3y.  Ans.     +3. 
3O.  Divide  +9y  by  —  3y.  Ans.    —  3. 


*  The  terra  quotient  would  be  more  exact  and  technical  here  ;  but,  in  re- 
sults hereafter,  we  shall  invariably  use  the  term  Ans.,  as  more  brief  and  ele- 
gant, and  it  is  equally  well  understood. 


DIVISION.  27 

(Art.  17.)  The  product  of  a3  into  «2  ;s  a5,  (Art.  10,)  that  is, 
in  multiplication  we  add  the  exponents  ;  and  as  division  is  the 
converse  of  multiplication,  to  divi  le  powers  of  tKe  same  letter, 
we  must  subtract  the  exponent  of  the  divisor  from  that  of  the 
dividend. 

Divide  2a°  by  a4.  Ans.     2a*, 

Divide  —  a7  by  a6.  Jlns.     —  a. 

Divide  IGr5  by  4x.  tfns.     4x*. 

Divide  ISaxy*  by  —  Say.  Jlns.     —  5xyz. 

Divide  63am  by  7an.  Aw.     9«"-n. 

Divide  I2axn  by  —  Sax.  JLns. 

Divide  7rf6  by 

Divide  —  5a2^  by  —  -7fl4a;2.  ^ns.     -^, 


Divide  117aW  by  78«56c4.  . 

o 

Divide  96a6c  by  \Za*bc*d.  Ans. 


Divide  a^c2  by 

n 

Divide  27a364cd2  by  21  abed.  Ans.    -d>b*d. 

Divide  14a62cc?  by  Ga^c2.  £ns.     —. 

ottC 

(Art.  18.)  The  object  of  this  article  is  to  explain  the  nature  of 
negative  exponents. 

Divide  a4  successively  by  a,  and  we  shall  have  the  following 
quotients  : 

1       1       1 
a3,    a2,    a,     1,    ~,    ^,    ^,     &c. 

Divide  fl4  again,  rigidly  adhering  to  the  principle  that  to 
divide  any  power  of  a  by  a,  the  exponent  becomes  one  less,  and 
we  have 


28  ELEMENTS   OF  ALGEUKA. 

a  ,     a2,     a1,     a°,     or1,     or2,     er3,     &c. 

Now  these  quotients  must  be  equal,  that  is,  a3  in  one  series 
equals  a3  in  the  other,  and 

rt2=a2,    a=al,     l=a°,    -=<*-1     —=a~*    —  =a~* 
a  a2  a3 


Another  illustration.  We  divide  exponential  quantities  by 
subtracting  the  exponent  of  the  divisor  from  the  exponent  of  the 
dividend.  Thus  a5  divided  by  a2  gives  a  quotient  of  «5~2=o3. 
a5  divided  by  a7=a5~7=a~2.  AVe  can  also  divide  by  taking  the 
dividend  for  a  numerator  and  the  divisor  for  a  denominator,  thus 

i5     I  1 

—  =—,  therefore  —  2=a~2  (Axiom  7.) 

a1    a2  a2 

From  this  we  learn,  that  exponential  factors  maybe  changed 
from  a  numerator  to  a  denominator^  and  the  reverse,  by  chang- 
ing the  signs  of  the  exponents. 


Thu<          -ax~*        —~  -=a-- 

US'     ^~  3*r'~3a3        xn 

Divide  d>bc  by  a*b2c~l.  Am.     arlb~lc*. 

Observe,  that  to  divide  is  to  subtract  the  exponents. 


Divide  Uab*cd  by  6a2bc*.  dns.  = 

(Art.  19.)  A  compound  quantity  divided  by  asimple  quantity,  is 
effected  by  dividing  each  term  of  the  compound  quantity  by  the 
simple  divisor. 

EXAMPLES. 

1.  Divide  Sax  —  15x  by  3x.  Ana.     a  —  5. 

».  Divide  8a^+12^  by  4x*.  Jlns.     2x+3. 

3.  Divide  3bcd+l2bcx—  962c  by  3bc.  Jlns.     d-\-4x—  3b. 

4.  Divide  lax-\-3ay  —  7bd  by  —  lad. 

x     3  y  ,  b 

tins.   -  T  —  r^H— 

d     Id    a 


DIVISION.  29 

ft.  Divide  15«26c  —  \5acx*-\-5ad*  by  —  5ac. 

a 
fins.     — 


c 
6.  Divide  Wx*—I5x2  —  25a?  by  5x.        Jlns.    2x*—  3x  —  5. 

V    Divide  —  -Wab+GQab*  by  —  Qab. 


.         —I  Oft9. 

o 


8.  Divide  36«262-f  60«2&—  6a&  by  —1206. 


O.  Divide    lOrx  —  cry-{-2crx  by  cr. 

10.  Divide    10uy-r-16d  by  2d. 

11.  Divide   6«?/  —  18aceH-24a   by   Ca. 

12.  Divide  mx  —  amx-\-m  by  m. 

(Art.  20.)  We  now  come  to  the  last  and  most  important  ope- 
ration in  division,  the  division  of  one  compound  quantity  by  an- 
other compound  quantity. 

The  dividend  may  be  considered  a  product  of  the  divisor  into 
the  yet  unknown  factor,  the  quotient  ;  and  the  highest  power  of 
any  letter  in  the  product,  or  the  now  called  dividend,  must  ba 
conceived  to  have  been  formed  by  the  highest  power  of  the  same 
letter  in  the  divisor  into  the  highest  power  of  that  letter  in  the 
quotient.  Therefore,  both  the  divisor  and  the  dividend  must 
be  arranged  according  to  the  regular  powers  of  some  letter. 

After  this,  the  truth  of  the  following  rule  will  become  obvious 
by  its  great  similarity  to  division  in  numbers. 

RULE.  Divide  thejirst  term  of  the  dividend  by  the  first  term 
of  the  divisor,  and  set  the  result  in  the  quotient.'* 

Multiply  the  whole  divisor  by  the  quotient  thus  found,  and 
subtract  the  product  from  the  dividend. 

The  remainder  will  form  a  new  dividend,  with  which  pro- 
ceed as  before,  till  the  first  term  of  the  divisor  is  no  longer 
contained  in  the  first  term  of  the  remainder. 

The  divisor  and  remainder,  if  there  be  a  remainder,  are  then 

*  Divide  the  first  term  of  the  dividend  and  of  the  remainders  by  the  first 
term  of  the  divisor  ;  be  nut  troubled  about  other  terms. 


30  ELEMENTS  OF  ALGEBRA. 

to  be  written  in  the  form  of  a  fraction,  as  in  division  of  num- 
bers. 

EXAMPLES. 

Divide   a*+2ab+&   by  a+b. 

Here,  a  is  the  leading  letter,  and  as  it  stands  first  in  both  the 
dividend  and  divisor,  no  change  of  place  is  necessary. 

OPERATION. 


ab 


ab+b* 
ab+b2 

Agreeably  to  the  rule,  we  consider  that  a  will  be  contained  in 
a2,  a  times  ;  then  the  product  of  a  into  the  divisor  is  a*-\-ab,  and 
the  first  term  of  the  remainder  is  ab,  in  which  a  is  contained  6 
times.  We  then  multiply  the  divisor  by  6,  and  there  being  no 
remainder,  a-}-b  is  the  whole  quotient. 

Divide   a3  +3  «*#-{-  Sao?  +#3  by  x+a. 

As  the  highest  power  of  a  stands  in  the  first  term  of  the  divi- 
dend, and  the  powers  of  a  decrease  in  regular  gradation  from 
term  to  term,  therefore  we  must  change  the  terms  of  the  divisor 
to  make  a  stand  first. 

OPERATION. 


a3-!-  azx 


2azx-\-2axz 


DIVISION.  8] 

a.  a— c)«3— 4a2c-|-4ac2— C3(fl2—3ac4-c« 


a3—  azc 


3 .  a2— 40-HK—  6a2-f- 1 2a— 8(0—2 


—  2a2-f-  8a—  -8 

4.  Divide   6a:4—  96   by   6z—  12.         .#n$.  a*-f  23*4-43+8. 

5.  Divide   a2  —  ft2  by  a  —  6.  ./#ns.     a+6. 

6.  Divide   25z6—  ^—  2x?—  8x2  by   5r»  —  4;r*. 


(Art.  21.)  We  may  cast  out  equal  factors  from  the  dividend 
and  divisor,  without  changing  the  value  of  the  quotient;  for 
amxy  divided  by  am  gives  xy  for  a  quotient  ;  cast  out  either  of 
the  common  factors  a  or  m  from  both  dividend  and  divisor,  and 
we  shall  still  have  xy  for  a  quotient.  This,  in  many  instances, 
will  greatly  facilitate  the  operation.  Thus,  in  the  4th  example, 
the  factor  6  may  be  cast  out,  as  it  is  contained  in  all  the  terms  ; 
and  in  the  6th  example  the  factor  a?2  may  be  cast  out  ;  the  quo- 
tients will  of  course  be  the  same. 

T.  Divide   a9-+4ax+4x*+y*  by   a+2x. 


8.  Divide   6a4-j-9a2  —  15a  by   So55  —  3a. 

(Observe  Art.  21.)  rfns.     2«8+2«+5. 

9.  Divide   a,-6—  y6  by  x*+2x*y+2xyz+y*. 

Ans.     x*—2x*y  -\-2xy*  —  y*. 


32  ELEMENTS  OF  ALGEBRA. 


10.  Divide  «£*—  («2-f  l>)tf+bz  by   ax—  b. 

tfns.     x  —  ax  —  b 

11.  Divide  1  by  1—  a.  J2ns.     1-f  a+a2-f  a8,  &c.,  &c 

12.  Divide   aM-£+?+l   by       +*. 


N.  B.  We  may  multiply  both    dividend   and  divisor  by  the 
same  number  as  well  as  divide  them. 


13.  Divide    1—  5*/-H<ty2—  Wy*+5y4—  y5  by    1—  27/-f-7/2. 

Am.     1—  3y4-3y2—  2/3 

14.  Divide    «4-f  4/;4   by   az—2ab+2b2. 

Am.     az-\-2ab-\-2bz. 

15.  Divide  a?6—  a?4-|-a-3—  ^2+2^~  1    by  a^+a?—  1. 

^/w.     a:4 

16.  Divide   a5  —  or5  by   a  —  #. 

Am. 

JT.  Divide  &5-f?/5   by 


18.  Divide   «3+5a2?/-j-5«?/2-{-i/3   by  a-fy. 

Ans. 

If  more  examples  are  desired  for  practice,  the  examples  in 
multiplication  may  be  taken.  The  product  or  answer  may  be 
taken  for  a  dividend,  and  either  one  of  the  factors  for  a  divisor  ; 
the  other  will  be  the  quotient. 

Also,  the  examples  in  division  may  be  changed  to  examples  in 
multiplication  ;  and  these  changes  may  serve  to  impress  on  the 
mind  of  the  pupil  the  close  connection  between  these  two  opera- 
tions. 

(Art.  22.)  In  the  following  examples  the  dividends  and  divi- 
sors are  given  in  the  form  of  fractions,  and  the  quotients  are  the 
terms  after  the  sign  of  equality.  Let  the  pupil  actually  divide, 
and  observe  the  quotients  attentively. 

a;2-a2 

1.  -  -  = 

x  —a 


DIVISION,  33 

2.  ^Zl^ 
x—  a 

3. 


4. 

a?  —  a 

Hence  we  may  conclude  that  in  general  xm  —  «m  is  divisible  by 
x  —  a,  m  being  any  entire  positive  number. 

That  is,  —  ~—=xm-l-\-axm-*Jr  -  -  -  arn^x-\-am~l 

The  quotient  commencing  with  a  power  of  x,  one  less  than 
m,  and  ending  with  a  power  of  a,  one  less  than  m. 

These  divisions  show,  that  the  difference  of  two  equal  powers 
of  different  quantities  is  always  divisible  by  the  difference  of 
their  roots. 

(Art.  23.)  By  trial,  that  is,  actual  division,  we  shall  find  that 

.r2—  az 

—  -;  —  =x  —  a. 
x-\-a 


x-}-a 

=  x5—  ax*+  aV 


x-\-a 

&c.     &c.     &c.     &c. 

From  which  we  learn  that  the  difference  of  any  two  equal 
powers  of  different  quantities,  is  also  divisible  by  the  sum  of  their 
roots  when  the  exponent  of  the  power  is  an  even  number. 

(Art.  24.)  By  actual  division  we  find  that 


--.  — 
x-\-a 


x-\-a 

And  in  general,  we  may  conclude  that  the  sum  of  any  two 
equal  powers  of  different  quantities,  is  divisible  by  the  sum  of 
their  roots  when  the  exponent  of  the  power  is  an  odd  number. 


34  ELEMENTS  OF  ALGEBRA. 

Li  Art.  22,  if  we  make  a=l  the  formulas  become 
x2—  I 


=aH-l. 


,  &c. 


8— 1 


a?  —1 
If  we  make  a?=l,  what  will  the  formulas  become  ? 

Make  the  same  substitutions  in  articles  23  and  24,  and  exa- 
mine the  results. 

By  inspecting  articles  22,  23,  and  24,  we  find  that 

=a?3—  as,   &c.. 


for  the  product  of  the  divisor  and  quotient  must  always  produce  the 
dividend.  These  principles  point  out  an  expedient  of  condensing 
a  multitude  of  terms  by  multiplying  them  by  the  roots  of  the  terms 
involved.  Thus,  a^iaa^+aVzta^+a4?  can  be  condensed  to 
two  terms  by  multiplying  them  by  #  ±a,  the  root  of  the  first  and 
last  term,  with  the  minus  sign  where  the  signs  are  plus  in"  the 
multiplicand,  and  with  the  plus  sign  where  the  signs  are  alter- 
nately plus  and  minus.  See  examples  in  Art.  22  and  24. 


ALGEBRAIC  FRACTIONS. 

(Art.  25.)  We  shall  be  very  brief  on  the  subject  of  algebraic 
fractions,  because  the  names  and  rules  of  operations  are  the  same 
as  numeral  fractions  in  common  arithmetic ;  and  for  illustration, 
shall,  in  some  cases,  place  them  side  by  side. 

CASE  i.  To  reduce  a  mixed  quantity  to  an  improper  frac- 
tion, multiply  the  integer  by  the  denominator  of  the  fraction^ 
and  to  the  product  add  the  numerator,  or  connect  it  with  its 
proper  sign,  -{-  or  — ;  then  the  denominator  being  set  under 
this  sum,  will  give  the  improper  Jraction  required. 


ALGEBRAIC  FRACTIONS.  35 


EXAMPLES. 

1.  Reduce  2f  and  G-TT  to  improper  fractions. 

Jhu.     y  and  *+* 

These  two  operations,  and  the  principle  that  governs  them, 
aie  exactly  alike. 

2.  Reduce  5J  and  «+y  to  improper  fractions. 

An,.     y  .ad  ?*+£*" 

3.  Reduce  4  —  |  and   a  --  to  improper  fractions. 

x 

,  ax  —  b 

MS.     y  and  -- 
x 

,    „   T  4  —  1  3#  —  « 

«.  Keduce  5  --  -  —  and  20  —  --  to  improper  fractions. 


5.  Reduce  5«-|-  —  y  —  to  an  improper  fraction. 

6.  Reduce    12-J  --  -r  —  to  an  improper  fraction. 

7.  Reduce  4-f-2a?H  —  to  an  improper  fraction. 

c 


8.  Reduce  5a;  -     -  to  an  improper  fraction. 

3 

9.  Reduce  3«  —  9  --  T-TT—  to  an  improper  fraction. 

- 


* 

a+3 

CASE  2.  TVte  converse  of  Case  1.  Tb  reduce  improper 
fractions  to  mixed  quantities,  divide  the  numerator  by  the  de- 
nominator, as  far  as  possible,  and  set  the  remainder,  (if  any,} 


36  ELEMENTS  OF  ALGEBRA. 

over  the  denominator  for  the  fractional  part ;  the  two  joined 
together  with  their  proper  sign,  will  be  the  mixed  quantity 
sought. 

EXAMPLES. 

1 .  Reduce  4¥7  and  — ^ —  to  mixed  quantities. 

Jins.     5J  and  0-f-y 

2.  Reduce  *•/  and  —      -  to  mixed  quantities. 

dns.     2f  and  0-f-- 

3.  Reduce  — to  a  mixed  quantity. 

ab+y 


50- 

y 

4.  Reduce  -    — — -  to  a  whole  or  mixed  quantity. 


Jlns. 


5.  Reduce —  to  a  whole  number. 

x—y 


Am.     2(s?+xy+y2)  by  (Art.  22.) 

6.  Reduce  -  to  a  mixed  quantity. 

c 

1002—  4a-j-6 

7.  Reduce  -  -  --  to  a  mixed  quantity. 

5a 


8.  Reduce  —  -  —  to  a  mixed  quantity. 

3 

3#2  —  I2ax-}-y  —  9x 

9.  Reduce  -  2  -   to  a  mixed  quantity. 

(Art.  26.)  It  is  very  desirable  to  obtain  algebraic  quantities  in 
their  most  condensed  form.  Therefore,  it  is  often  necessary  to 
reduce  fractions  to  their  lowest  terms  ;  and  this  can  be  done  as 
in  arithmetic,  by  dividing  both  numerator  and  denominator  by 


ALGEBRAIC  FRACTIONS.  37 

their  obvious  common  factors,  or  for  their  final  reduction,  by 
their  greatest  common  measure.  If  the  terms  have  no  common 
measure,  the  fraction  is  already  to  its  lowest  terms. 

The  principle  on  which  these  reductions  rest  is  that  of  divi- 
sion, explained  in  (Art.  21). 

CASE  3.  To  find  the  greatest  common  measure  of  the  terms 
of  a  fraction,  divide  the  greater  term  by  the  less,  and  the  last 
divisor  by  the  remainder,  and  so  on  till  nothing  remains  ;  then 
the  divisor  last  used  will  be  the  common  measure  required. 

But  note,  that  it  is  proper  to  arrange  the  quantities  according 
to  the  powers  of  some  letter,  as  is  shown  in  division. 

N.  B.  During  the  operation  we  may  cast  out,  or  throw  in  a 
factor  to  either  one  of  the  terms  which  is  not  a  factor  in  the 
other,  as  such  a  factor  would  make  no  part  of  the  common 
measure,  and  the  value  of  quantities  is  not  under  consideration. 

Thus,  the  fraction  — - — —  has  a4-b  for   its   greatest  common 
a2 — b* 

measure;  and  this  quantity  is  not  affected  by  casting  out  the 
factor  b  from  the  numerator,  and  seeking  the  common  measure 

a+b 

of  the  fraction  — — — . 
a2 — b2 

(Art.  27.)  To  demonstrate  the  truth  of  the  rule  for  finding  the 
greatest  common  measure,  let  us  suppose  D  to  represent  a  divi- 
dend, and  d  a  divisor,  q  the  first  quotient  and  r  the  first  re- 
mainder. 

Tn  short,  let  us  represent  successive  divisions  as  follows  : 

d)D(q 
dq 


0 

Now,  in  division,  the  dividend  is  always  equal  to  the  product 
of  the  divisor  and  quotient,  plus  the  remainder,  if  any. 


38  ELEMENTS  OF  ALGEBRA. 


Therefore,  r=r'q" 

and  d=rq'-}-r' 

and  D=dq  -f-r. 

As  r^=r'q",  the  last  divisor  r'  is  a  factor  in  r  (there  being  no 
remainder)  ;  that  is,  r'  measures  r. 

Now  as  r'  measures  r,  it  measures  any  number  of  times  r, 
or  r<7'-f-r',  orrf;  therefore  r'  measures  d. 

Again,  as  r'  measures  d  and  r,  it  measures  any  number  of 
times  d  -\-r  ;  that  is,  it  measures  dq-\-r  or  D. 

Hence  r',  the  last  divisor,  is  a  common  measure  to  both  D 

and  d,  or  of  the  terms  of  the  fraction  __ 

d 

We  have  now  to  show  that  r'  is  not  only  the  common  mea- 
sure of  D  and  d,  but  the  greatest  common  measure. 

In  division,  if  we  subtract  the  product  of  the  divisor  and  quo- 
tient from  the  dividend,  we  shall  have  the  remainder. 

That  is,  D  —  dq=r,  and  d—rq'—r'. 

Now,  every  common  measure  of  D  and  d  is  also  a  measure 
of  r,  because  D  —  dq=r  ;  for  the  same  reason  every  common 
measure  of  r  and  d  is  also  a  measure  of  r'.  But  the  greatest 
measure  of  r'  is  itself.  This  final  remainder  is,  therefore,  the 
greatest  common  measure  of  D  and  d. 

EXAMPLES. 
1.  Find  the  greatest  common  measure  of  the  two  terms  of 

the  fraction  —  -  —  -  and  with  it,  reduce  the  fraction  to  its  lowest 


terms. 

CONSIDERATION    AND    OPERATION. 

The  denominator  has  a3  as  a  factor  to  all  its  terms,  which  is 
not  a  factor  in  the  numerator  ;  hence  this  can  form  no  part  of 
the  common  measure,  or  the  common  measure  will  still  be  there 
if  this  factor  is  taken  away. 

We  then  seek  the  common  measure  of  a4  —  1  and  a?-\-l. 


ALGEBRAIC  FRACTIONS. 


—a2—! 

Hence  a2~\-l  is  the  common  measure,  which,  used  as  a  divisor 
to  both  numerator  and  denominator,  reduces  the  fraction  to 
a2—! 

"  a3    ' 

2.  Find  the  greatest  common  measure,  and  reduce  the  frac- 
tion 


Divide  this  rem.  by  y2  zy2 — ?/3 


ay— 2/2 
*y— af 


3.  Find  the  greatest  common  measure  and  reduce  the  frac- 
uon 


Common  measure    #  —  ^y. 
Fraction  reduced 


Ans.     Greatest   common  measure  «2+a^y.     Reduced  frac- 

.ion  *±& 
5axy. 


40  ELEMENTS  OF  ALGEBRA. 


Find   the   greatest   common   measure  of 
and  a2c+2abc+b2c. 

Reject  the  common  factor  c  in  one  of  the  quantities, 


a'-f  2azb+  ab* 


4.  Find  the  greatest  common  divisor,  and  reduce  the  fraction 
to  its  lowest  terms. 


Here  we  find  that  neither  term  is  divisible  by  the  other  ;  but 
if  these  quantities  have  a  common  divisor,  such  divisor  will  still 
exist  if  we  multiply  one  of  the  terms  by  any  number  whatever, 
to  render  division  possible. 

Therefore  take  4a3  —  2a2  —  3a-\-  I 
Multiply  it  by   3 


2__2a—  l)12a3—  6a2—  9a+  3(4a 
12a3—  8a2—  4a 


2a2—  5a+  3 
Multiply  by  3 


9(2 
—  4a—  2 


Divide  by —11  —  lla-j-11 


a— 1 
a— 1 

Jins      Greatest  common  measure  a— 1.     Reduced  fraction 

j*<zi:L_ 

4^+20—1* 


ALGEBRAIC  FRACTIONS.  41 

5.  Find  the  greatest  common  measure  and  reduce  the  frac- 

to  its  lowest  terms' 


Jlns.     Common  divisor  a2 — x2.     Reduced  fraction . 

a — x 

<*•  Find  the  greatest  common  measure,  and  reduce  the  frac- 
tion — to  its  lowest  terms. 


Jlns.     Common  divisor  a? — y2.     Red.  frac. 


.  Reduce 


g.  Reduce 


, 
a2  —  xz 


to  its  lowest  terms. 
tins. 


a+x 

to  its  lowest  terms.  ^  & 

3a — x 


Ans. 


9.  Reduce 

10.  Reduce 


—  -  to  its  lowest  terms. 
—  So3 


Am. 


bx 


to  its  lowest  terms. 


— 
11.  What  is  the  value  of  -- 


Ans. 


12.  Find  the  greatest  common  divisor  of  12a4  —  24a36+  12a262, 
and  8o362- 


(A  rt.  28.)  We  may  often  reduce  a  fraction  by  separating  both 
numerator  and  denominator  into  obvious   factors,  without   the 
formality  of  finding  the  greatest  common  divisor.     The  follow 
ing  are  some  examples  of  the  kind  : 
4 


42  ELEMENTS  OP  ALGEBRA. 

1.  Reduce  —  --        -  to  its  lowest  terms. 


a(a2—b2)     _a(a—b}(a+b} 


2.  Reduce  —  -  —  —  to  its  lowest  terms.  Jlns. 


-  T—      IV    AfcO    lUWtOl     1C  HHO.  VJ.IIO.      — -      -      — « 

^2 J  ,p | 

3.  Reduce ; —  to  its  lowest  terms.  Jlns.     . 

zy+y  y 

CX-}- CX2  C-}-CX 

4.  Reduce to  its  lowest  terms.          Jins.    — . 

acx-{-abx  ac-\-ab 

2#3— .I6x— 6 

5.  Reduce  — to  its  lowest  terms.  Jlns.  I. 

3^3 — 24x — 9 

(Art.  29.)  To  find  the  least  common  multiple  of  two  or 
more  quantities. 

The  least  common  multiple  of  several  quantities  is  the  least 
quantity  in  which  each  of  them  is  contained  without  a  remainder. 

Thus,  the  least  common  multiple  of  the  prime  factors,  a,  b,  c, 
x,  is  obviously  their  product  abcx.  Now  observe  that  the  same 
product  is  the  least  common  multiple  also,  when  either  one  of 
these  letters  appears  in  more  than  one  of  the  terms.  Take  a, 
for  example,  and  let  it  appear  with  b,  c,  or  xf  or  with  all  of 
them,  as  a,  ab,  c,  ax,  or  a,  6,  «c,  ax,  the  product  abcx  is  still 
divisible  by  each  quantity.  Therefore,  when  the  same  factor 
appears  in  any  number  of  the  terms,  it  is  only  necessary  that  it 
should  appear  once  in  the  product;  that  is,  once  in  the  least 
common  multiple.  If  it  should  be  used  more  than  once,  the 
product  so  formed  would  not  be  the  least  common  multiple. 

From  this  examination,  the  following  rule  for  finding  the  taast 
common  multiple  will  be  obvious : 

RULE,  Write  the  given  quantities,  one  after  another,  and 
draw  a  line  beneath  them.  Then  divide  by  any  prime  factor 
that  will  divide  two  or  more  of  them  without  a  remainder, 
bringing  down  the  quotients  and  the  quantities  not  divisible, 
to  a  line  below.  Divide  this  second  line  as  the  first,  forming 


ALGEBRAIC  FRACTIONS.  43 

a  third,  fyc., until  nothing  but  prime  quantities  are  left.     Then 
multiply  all  the  divisors  and  the  remainders  that  are  not  divis- 
ible, and  their  product  will  be  the  least  common  multiple. 
N.  B.  This  rule  is  also  in  common  arithmetic. 

EXAMPLES.  I 

1.  Required  the  least  common  multiple  of  Sac,  4a2, 
8«c,  and  ex. 

2a)Sac        4a2         I2ab        Sac        ex 


2c)4c 

2a 

Qb 

4c 

ex 

2)2 

a 

3b 

2 

X 

1 

a 

36 

1 

X 

Therefore  2a  X  2c  X  2  X  a  X  36  X  x=24a2cbx. 

Here  the  divisor  2c  will  not  divide  2a,  but  the  coefficient  of 
c  will  divide  the  coefficient  of  a,  and  we  let  them  divide,  for  it 
is  the  same  as  first  dividing  by  2,  and  afterwards  by  c.  From 
the  same  consideration  we  permit  2c  to  divide  ex,  or  let  the  let- 
ter c  in  the  divisor  strike  out  c  before  x. 

By  the  rule  we  should  divide  by  2  and  by  c  separately ;  but 
this  is  a  practical  abbreviation  of  the  rule. 

2.  Required  the  least  common  multiple  of  27 a,  156,  9a6,  and 
3a2.  Ans.     135a26. 

3.  Find  the   least  common  multiple  of  (a2 — x2},   4(a — x), 
and  (a+x).  tins.     4(a2— #*). 

£•  Find  the  least  common  multiple  of  ax2,  bx,  acx,  and 
a2— x2.  Am.  (a?—a?)acbx?. 

5.  Find  the  least  common  multiple  of  a-f-6,  a — b,  a2-{-ab-{-b2, 
and  a2— ab+b2.  Ans.  a6— b6. 

The  least  common  multiple  is  useful  many  times  in  reducing 
fractions  to  their  least  common  denominator. 

CASE  4.     To  reduce  fractions  to  a  common  denominator. 

(Art.  30.)  The  rule  for  this  operation,  and  the  principle  on 
which  it  is  founded,  is  just  the  same  as  in  common  arithmetic, 
merely  the  multiplication  of  numerator  and  denominator  by  the 


44  ELEMENTS   OF  ALGEBKA. 

same  quantity.  The  object  of  reducing  fractions  to  a  common 
denominator  is  to  add  them,  or  to  take  their  difference,  as  diffe- 
rent denominations  cannot  be  put  into  one  sum. 

RULE.  Multiply  each  numerator  by  all  the  denominators, 
except  its  own,  for  a  new  numerator,  and  all  the  denominators 
for  a  common  denominator. 

Or,  find  the,  least  common  multiple  of  the  given  denomina- 
tors for  a  common  denominator;  then  multiply  each  denomi- 
nator by  such  a  quantity  as  will  give  the  common  denomina- 
tor, and  multiply  each  numerator  by  the  same  quantity  by 
which  its  denominator  was  multiplied. 

EXAMPLES. 

1.  Reduce  —  and  —  to  a  common  denominator. 
x          2c 

4ac         3bx 

rfns.     -  and  -  . 

2cx        2cx 

2a        3a-f-26 

8.  Reduce  —  and  -  to  a  common  denominator. 
b  2c 

4ac       ,  Bab+2  I? 
Jlns.     ——  and  --- 
2bc  2bc 

3.  Reduce  —  and  —  ,  and  4d  to  a  common  denominator. 
3x         2c 


,  9bx      ,  24cdx 

Jins.     -  and  —  and  -  . 
Gcx         Gcx          Gcx 

4.  Reduce  7-,  -  ,  ~  —  ,  to  fractions  having  a  common  de- 
0       c       x-\-a 

Jlns.     acx-{-a2c     (bx-\-b)(x-}-a)         bey 

bcx  -\-abc        b  ex  -\-abc     '    bcx-\-abc' 

(Art.  31.)     CASE  5  Addition  or  finding  the  sum  of  fractions. 

RULE.  Reduce  the  fractions  to  a  common  denominator  ; 
and  the  sum  of  the  numerators,  written  over  the  common  deno- 
minator, will  be  the  sum  of  the  fractions. 


ALGEBRAIC  FRACTIONS.  45 


EXAMPLES. 


3x    2x  X 

1    Add  —  ,  —  ,  and  -  together 
57  * 


I28.r 

//  *  j  o  -.  ___     _.  _  _  .^   -_-,-._  —  —  >   _ 

105  105 


,    a        .    a-\-b 
2.  Add  T   and   —  ~.  Ans. 


T  —  ~.  .     -  - 

bc  be 


3.  Add  -,  -  and   -—  . 


f±»   and  . 

a  —  6  a  4-  6 

a-f-3  2a—  5  , 

5.  Add  2a+-     -   and   4aH  ---  .  .^/?5.  6«H  —  077- 


8x*  %ax  %abx — 8cz* 

6.  Add  a — — r-    and  b-{ Jins.    a-\-o-\ •— . 

b  c  uc 

x — 2  2;r — 3 

7.  Add  5.r+— --   and    4x — . 

o  ox 

Ans. 


3.r       b  ,    6  —  a?. 

8.  What  is  the  sum  of  26-{—  -,  7  -   and   —  —  -  ? 

5    6  —  x  b 


2  —  5bx 


Mt  ___    f)  Q,j  •         O 

9    What  is  the  sum  of  5y-l-^—  -    and   4y  --  «C  1  ? 

d  o 


4C  ELEMENTS  OF  ALGEBRA. 

1.0.   What  is  the  sum  of  50,  —  ,  and  ^~ 


Ans. 

(Art.  32.)     CASE  G.     Subtraction  or  finding  difference. 

RULE.  Reduce  the  fractions  to  a  common  denominator,  and 
subtract  the  numerator  of  that  fraction  which  is  to  be  sub- 
tracted from  the  numerator  of  the  other,  placing  the  difference 
over  the  common  denominator. 

EXAMPLES. 

7x  2x—  1  2lx—Ax-\-2     l7x-\-2 

1.  From  -take-—  .         An*.     -  -  -  =__. 


take  .          Eq.  fractions  — — -2» 

Difference  or  *ftns. 

nt  O/y«  rp 

3.  From  5  take  ~.  Diff.     ^ 

o  7  «1 

4.  From  ^—  take  — .  Ans.     — — 

7  y  63 

2a— b  Ba — 4b 

9    From  — ; —   subtract  — -7 — . 
4c  3o 


126c 

.  llfl — 10  ,  3a — 5. 

6.  From  3a-\ subtract    2a-\ . 

15  7 

•tins,     a- 


.  From  x-\  —  ~~*    subtract  —  —  ^—.          fins,     x 

x2—xy 


a—b      .      2b—4a  5ad  —  5bd—  4bc-\-Sac 

8.  From  -—  ~  take  —  —  —  .      Jim.  --  - 
2c  bd  IQcd 


ALGEBRAIC  FRACTIONS.  47 

.  x  x — a  .  cx-\-bx~ ab 

9.  From  3x-\--r  take  x .        fins.  2x-{ -, 

be  be 

a-\-b  a — b 

10.  Find  the  difference  between   -  — r    and    — J-T-. 

a — b  a+b 

4ab 
tins. 


a2—  ft2 


11.  prom  take  .  fa.     4. 

xy  xy 

CASE  7.     Multiplication  of  fractions. 

(Art.  33.)  The  multiplication  of  algebraic  fractions  is  just  the 
fame  in  principle  and  in  fact,  as  in  numeral  fractions,  hence  the 
rule  must  be  the  same. 

It  is  perfectly  obvious,  that  f  multiplied  by  2  must  be  -|,  and 
multiplied  by  3  must  be  f  ;  and  the  result  would  be  equally  ob- 
vious with  any  other  simple  fraction  ;  hence,  to  multiply  a  frac- 
tion by  a  whole  number,  we  must  multiply  its  numerator. 

It  is  manifest  that  doubling  a  denominator  without  changing  its 
numerator  halves  a  fraction,  thus  £  ;  double  the  2,  and  we  have 
?,  the  half  of  the  first  fraction. 

Also  f  ,  double  the  5  gives  T\,  the  half  of  |.  In  the  same 
manner,  to  divide  a  fraction  by  3  we  would  multiply  its  denomi- 
nator by  3,  &c.  In  general,  to  divide  a  fraction  by  any  num- 
ber, we  must  multiply  the  denominator  by  that  number. 

Now  let  us  take  the  literal  fraction  -r,  and  multiply  it  by  c,  the 

ac 
product  must  be  -j-  . 

Again,  let  it  be  required  to  multiply  -  by  -.     Here  the  mul- 

tiplication is  the  same  as  before,  except  the  multiplier  c  is  divided 
by  d  ;  therefore  if  we  multiply  by  c  we  must  divide  by  d.     But 

the  product  of  -  by  c  is  —  ;  this  must  be  divided  by  d1  and  we 
ac 


shall  have  -— ;  for  the  true  product  of  ,-  by  -=. 
bd  b     J   d 


48  ELEMENTS  OF  ALMEBliA. 

From  the  preceding  investigation  we  draw  the  following  rule 
to  multiply  fractions : 

RULE.  Multiply  the  numerators  together  for  a  new  nume- 
rator, and  the  denominators  together  for  a  new  denominator. 

N.  B.  When  equal  factors,  whether  numeral  or  literal, 
appear  in  numerators  and  denominators,  they  may  be  canceled, 
or  left  out,  which  will  save  subsequent  reductions. 

EXAMPLES. 

1.  Multiply  £  by  *  and  ®.  Al».     a^L. 

7   b     J  x         c  ex 

In  this  example,  b  in  the  denominator  of  one  fraction  cancels 
0  in  the  numerator  of  another. 


«•   iviuiiipiy 

< 

3.  Multiply  : 
4U  Multiply 
5.  Multiply* 

fi         TVTiiltiT-wlir 

f\f\              U  '  J       f\  f         1           \ 

ou            o{a~\~x\ 

**+%  bv  2« 

18 

Ans.     -. 
5x 

(a-x)a 

«2  —  x2  ,        2a 

V»tr 

2i/            a-j-a? 
ar2  —  i/2       3?               a 

2                       onrl 

y 

Ans.    a. 

•,             ana 
a;         07+y         a;—  y 

a?+l       ,  ^  —  1 

N.  B.  Reduce  mixed  quantities  to  improper  fractions. 

7.  What  is  the  continued  product  of  —  —  ,  -  :     :    and 

a-\-b     ax-\-x* 

ax  «2(rt—  b) 


a—x  x 

4v*  15|/  —  30 

8.  Multiply  ^-^    by     -*—  .  AM. 

*  Separate  into  factors  when  separation  is  obvious. 


ALGEBRAIC  FRACTIONS.  49 


9.  Multiply  -  —  by  -7—75.  •#»*• 

a-\-b         ab  —  b*  b 

10.  Multiply  -^P-,  by  jgg^gj*  ^n5'     3(«+:F)- 
11    Required  the  continued  product  of  -=  —  r-,  -^-T—  j    and 

w  ~~~O       flr~~it 


.5n*.     («+*). 


19.  Multiply  «+      by  «-.      .*».. 


_       . 
14.  Mult,ply  -—  —   by 


4«2—  1662  ,  56 

15.  MulUply  ---  by 


8a.+32afe+32ft. 


CASE  8      Division  of  Fractions. 

(Art.  34.)  To  acquire  a  clear  understanding  of  division  in 
fractions,  let  us  return  to  division  in  whole  numbers. 

The  first  principle  to  which  we  wish  to  call  the  attention  of 
the  reader,  is,  that  if  we  multiply  or  divide  both  dividend  and 
divisor  of  any  sum  in  division,  by  any  number  whatever,  we  do 
not  affect  or  change  the  quotient.  (Art.  21.) 

Thus,  2)6(3  4)12(3  8)24(3          &c 

The  second  principle  to  which  we  would  call  observation  is, 
that  if  we  multiply  any  fraction  by  its  denominator,  we  have  the 
numerator  for  a  product. 

Thus,  5  multiplied  by  3  gives  1,  the  numerator,  and  |  by  5 

gives  2,  and  -r  multiplied  by  b  gives  a,  <fec. 
5 


50  ELEMENTS  OF  ALGEBRA. 

Now  let  it  be  required  to  divide  j-  by  -. 

The  quotient  will  be  the  same  if  we  multiply  both  dividend 
and  divisor  by  the  same  quantity.  Let  us  multiply  both  terms 

by  «?,  the  denominator  of  the  divisor,  and  we  have  -=-  to   be  di 

vided  by  the  whole  number  c.     But  to  divide  a  fraction  by  a 
whole  number  we  must  multiply  the  denominator  by  that  num- 

ber.    (Art.  33.)     Hence  j~  is  the  true  quotient  required. 

We  can  mechanically  arrive  at  the  same  result  by  inverting 
the  terms  of  the  divisor,  and  then  multiplying  the  upper  terms 
together  for  a  numerator,  and  the  lower  terms  for  a  denominator; 
therefore  to  divide  one  fraction  by  another  we  have  the  following 

RULE.  Invert  the  terms  of  the  divisor,  and  proceed  as  in 
multiplication. 

EXAMPLES. 

a+b   .         c 

1.  Divide  —  —    by   —  p-r.  rfns. 

c  a-\-b 

2.  Divide  5X   by   -. 

a  c 

c     5x    5  ex 

Operation:  divisor  inverted  -rX  —  =  —  r- 

b      a       ab 


10«c* 

3.  Divide  -    by   —„  --  ,. 

a  —  x          a2  —  & 

I5ab     (a-\-x}(a—  x)  3b(a+x) 

Operation,  -  X^—  -  —  -  --  '-.     ^ns.  —  ^--—  -. 
a  —  x  lOac  2c 

2*2—  7  as 

4.  Divide  -  —  by  —  -—-s.      Ans. 

x-\-a      J  x*  -\-1ax-\-  a* 


Operation, 


Divide  into  factors,  in  all  such  cases,  and  cancel. 


ALGEBRAIC  FRACTIONS. 


51 


a2-\-ax-}-x2' 
70#  —  15 

5          7      25     * 

/?«?         ~~ 

5                 5 

x 

18x—  21 
Jlns.    —  —  ^  —  r  —  . 

'    x+l      J      3    ' 
1O.   Divide                 hv 

3a2 


,l.Dmde____+_by 


Am. 


12.  Divide  by 

5  5 


,_....     na — nx  .     ma — mx 

13.  Divide  TT-  by i~7— . 

- 


Ans.    — 
m 


14.  Divide  12  by 

15.  Divide 


,      a 
by  -. 
x          J  x 


,.,    T,.  .,     x — b         3cx 

lo.  Divide   — 5—   by  — r. 

J 


Jlns. 


I2x 


a*-fax-\-x* 

Jlns.    b+- 
a 


Divide  a  by  the  product  of  — : —  into   . 

X+y  x—y 


Jlns. 


18.  Divide  ^-i-rr  by  the  product  of  ^-^;  into   ~. 
ofx._i_A\      J  2a  «4-6 


52  ELEMENTS  OF  ALGEBRA. 

SECTION    II. 
CHAPTER  I. 

Preparatory  to  the  solution  of  problems,  and  to  extended  in- 
vestigations of  scientific  truth,  we  commenced  by  explaining  the 
reason  and  the  manner  of  adding,  subtracting,  multiplying,  and 
dividing  algebraic  quantities,  both  whole  and  fractional,  that  the 
mind  of  the  pupil  need  not  be  called  away  to  the  art  of  perform- 
ing these  operations,  when  all  his  attention  may  be  required  on 
the  nature  and  philosophy  of  the  problem  itself. 

For  this  reason  we  did  not  commence  with  problems. 

Analytical  investigations  are  mostly  carried  on  by  means 

OF  EQUATIONS. 

(Art.  35.)  An  equation  is  an  algebraical  expression,  meaning 
that  certain  quantities  are  equal  to  certain  other  quantities.  Thus, 
3-f-4=7  ;  a-{-6=c  ;  #4-4=10,  are  equations,  and  express  that 
3  added  to  4  is  equal  to  7,  and  in  the  second  equation  that  a 
added  to  b  is  equal  to  e,  <fcc.  The  signs  are  only  abbreviations 
for  words. 

The  quantities  on  each  side  of  the  sign  of  equality  are  called 
members.  Those  on  the  left  of  the  sign  form  the  first  member, 
those  on  the  right  the  second. 

In  the  solution  of  problems  every  equation  is  supposed  to  con- 
tain at  least  one  unknown  quantity,  and  the  solution  of  an  equa- 
tion is  the  art  of  changing  and  operating  on  the  terms  by  means 
of  addition,  subtraction,  multiplication,  or  division,  or  by  all  these 
combined,  so  that  the  unknown  term  may  stand  alone  as  one 
member  of  the  equation,  equal  to  known  terms  in  the  other 
member,  by  which  it  then  becomes  known. 

Equations  are  of  the  first,  second,  third,  or  fourth  degree, 
according  as  the  unknown  quantity  which  they  contain  is  of  the 
first,  second,  third,  or  fourth  power. 

ax-\-b=3az  is  an  equation  of  the  first  degree  or  simple 
equation* 


EQUATIONS.  53 

axz-\-bx=3ab  is  an  equation  of  the  second  degree  or  quadratic 
equation. 

axs-\-bx2-\-cx=2a4b  is  an  equation  of  the  third  degree. 
ax*-{-bx3Jrcx2-\-dx=2ab5  is  an  equation  of  the  fourth  degree. 

We  shall  at  present  confine  ourselves  to  simple  equations. 

(Art.  36.)  The  unknown  quantity  of  an  equation  may  be 
united  to  known  quantities,  in  four  different  ways  :  by  addition, 
by  subtraction,  by  multiplication,  and  by  division,  and  further 
by  various  combinations  of  these  four  ways  as  shown  by  the 
following  equations,  both  numeral  and  literal : 

NUMERAL.  LITERAL. 

1st.  By  addition,  aH-6=10  x-\-a=b 

2d.    By  subtraction,  x — 8  =  12  x — c=d 

3d.    By  multiplication,  20#=80  ax—e 

4th.  By  division,  7~16  ~d=£~^~a' 

5th.  ar+6— 8+4=10+2— 3,  x+a— b+c=d+c,  &c., 
are  equations  in  which  the  unknown  is  connected  with  known 
quantities  by  both  addition  and  subtraction. 

y*  or 

2aH-r=21>     ax-\—~Cj  are  equations  in  which  the  unknown 

o  0 

is  connected  with  known  quantities  by  both  multiplication  and 
division. 

Equations  often  occur,  in  solving  problems,  in  which  all  of 
these  operations  are  combined. 

(Art.  37.)  Let  us  now  examine  how  the  unknown  quantity  can 
be  separated  from  others,  and  be  made  to  stand  by  itself. 
Take  the  1st  equation,  or  other  similar  ones. 

x-f6=10  x-\-a=b 

Take  equal  quantities  6=  6  a=a     from  both 

members,  and  #=10 — 6  x=b — a         the 

remainders  must  be  equal.  (Ax.  2.)  Now  we  find  the  term 
sdded  to  x,  whatever  it  may  be,  appears  on  the  other  side  with 
a  contrary  sign,  and  the  unknown  term  x  being  equal  to  known 
terms  is  now  known. 


64  ELEMENTS  OF  ALGEBRA. 

Take  the  equations 
Add  equals  to  both  memb. 
Sums  are  equal  #=124-  8  x=d-{-c  (Ax.  1.) 

Here  again  the  quantity  united  to  x  appears  on  the  opposite 
side  with  a  contrary  sign. 

From  this  we  may  draw  the  following  principle  cr  rule  3f 
operation : 

Any  term  may  be  transposed  from  one  member  of  an  equa- 
tion to  the  other,  by  changing  its  sign. 

Now  20#=80.  ax—e.  If  we  divide  both  members  by  the 
coefficient  of  the  unknown  term,  the  quotients  will  be  equal. 

(Ax.  4.)     Hence  #=f  J=4.     x=-. 

That  is,  the  unknown  quantity  is  disengaged  from  known 
quantities,  in  this  case,  by  division. 

7*  "7* 

Again,  take  the  equations  -=16;  ~=g-\-a. 

Multiply  both  members  by  the  divisor  of  the  unknown  term, 
and  we  have  o?=16X4.  x=gd-\-ad.  Equations  which  must 
be  true  by  (Ax.  3.),  and  here  it  will  be  observed  that  x  is  libe- 
rated by  multiplication. 

From  these  observations  we  deduce  this  general  principle : 

That  to  separate  the  unknown  quantity  from  additional 
terms  we  must  use  subtraction;  from  subtracted  terms  we 
must  use  addition  ;  from  multiplied  terms  we  must  use  divi- 
sion ;  from  divisors  we  must  use  multiplication. 

In  all  cases  take  the  opposite  operation. 

EXAMPLES. 

1.  Given  3x — 4=7# — 16  to  find  the  value  of  #.     Ans.  #=3. 

2.  Given  3#-}-9 — 1 — 5x=0  to  find  the  value  of  x. 

Am.     x—±. 

3.  Given  4y-}-7—y-\-2l — 3-\-y  to  find  y.         Ans.  y=5h- 

4.  Given  Sax — c=b — 3ax  to  find  the  value  of  x. 

Jlns.    *=*-±-C 
8a 


EQUATIONS.  55 

5.  Given  axz-{-bx=9xz-{-cx  to  find  the  value  of  x  in  terms 

of  a,  b,  and  c.  jlns.     x= -. 

a — 9 

N.  B.  In  this  last  example  we  observe  that  every  term  of  the 
equation  contains  at  least  one  factor  of  x ;  we  therefore  divide 
every  term  by  x,  to  suppress  this  factor. 

(Art.  38.)  In  many  problems,  the  unknown  quantity  is 
often  combined  with  known  quantities,  not  merely  in  a  simple 
manner,  but  under  various  fractional  and  compound  forms. — 
Hence,  rules  can  only  embody  general  principles,  and  skill  and 
tact  must  be  acquired  by  close  attention  and  practical  application  : 
but  from  the  foregoing  principles  we  draw  the  following 

GENERAL  RULE.  Connect  and  unite  as  much  as  possible  all 
the  terms  of  a  similar  kind  on  both  sides  of  the  equation.  Then, 
to  clear  of  fractions,  multiply  both  sides  by  the  denominators, 
one  after  another,  in  succession.  Or,  multiply  by  their  con- 
tinued  product,  or  by  their  least  common  multiple,  (when  such 
a  number  is  obvious,)  and  the  equation  will  be  free  of  fractions. 

Then,  transpose  the  unknown  terms  to  the  first  member  of 
the  equation,  and  the  known  terms  to  the  other.  Then  unite 
the  similar  terms,  and  divide  by  the  coefficient  of  the  unknown 
term,  and  the  equation  is  solved. 

EXAMPLES. 

1.  Given    x-K#-{-3 — 7=6—1,    to   find   the   value   of  x. 
Uniting  the  known  terms,  after  transposition,  agreeably  to  the 
rule  of  addition,  we  find  x-}-zX=9.     Multiply  every  term  by 
2,  and  we  have   2x-\-x=lS.     Therefore   x=6. 

2.  Given   2x-{-$x-{-ix — 3a=46-f  30,    to  find   x. 

N.  B.  We  may  clear  of  fractions,  in  the  first  place,  before  we 
condense  and  unite  terms,  if  more  convenient,  and  among  literal 
quantities  this  is  generally  preferable. 

In  the  present  case  let  us  multiply  every  term  of  the  equation 
by  12,  the  product  of  3X4,  and  we  shall  have 

24;r-f9a:-|-4tf— 360=486+ 36a. 
Transpose  and  unite,  and  37#=48&-f-72a 


56  ELEMENTS  OF  ALGEBRA. 

Divide  by  37,  Md  *= 

3.  Given   5r-f-£#-f-*#=39,   to  find  the  value  of  x. 

Here  are  no  scattering  terms  to  collect,  and  clearing  of  frac- 
tions is  the  first  operation. 

By  an  examination  of  the  denominators,  12  is  obviously  their 
least  common  multiple,  therefore  multiply  by  12.  Say  12 
halves  are  6  whole  ones,  12  thirds  are  4,  12  fourths  are  3,  &c. 

Hence,  6x+4x+  3#=39  X  12 

Collect  the  terras,  13a?=39X  12 

Divide  by  13,  and  a?=  3X12=36,   Am. 

N.  B.  In  other  books  we  find  the  numerals  actually  multiplied 
by  12.  Here  it  is  only  indicated,  which  is  all  that  is  necessary. 
For  when  we  come  to  divide  by  the  coefficient  of  x,  we  shall 
find  factors  that  will  cancel,  unless  that  coefficient  is  prime  to  all 
the  other  numbers  used,  which,  in  practice,  is  very  rarely  the 
case. 

4.  Given   £a?-hia?-H#=0,   to  find  x. 

This  example  ia  essentially  the  same  as  the  last.  It  is  identi- 
cal if  we  suppose  a=39. 


Solution, 

Or,  13#=12a 

12a 

Divide  and  X=-T^ 

L*i 

Now  if  a  be   any  multiple  of  13,  the  problem  is  easy  and 
brief  in  numerals. 


,  3x—  11     5x—  5  ,  97—  7# 
5.  Given  2H  --  —  —  =  —  -  —  +  —  -  —  to  find  the  value  of  x. 

16  o  Z 

Here  16  is  obviously  the  least  common  multiple  of  the  deno- 
minators, and  the  rule  would  require  us  to  multiply  by  it,  and 
such  an  operation  would  be  correct  ;  but  in  this  case  it  is  more 
easy  to  multiply  by  the  least  denominator  2,  and  then  condense 
like  terms.  Thus, 


,     EQUATIONS.  57 

Multiply  by  2,  and  we  have 


Recollect  that  we  can  multiply  a  fraction  by  dividing  its  deno- 
minator. Also  observe  that  we  can  mentally  take  away  42  from 
both  sides  of  the  equation,  and  the  remainders  will  be  equal. 

(Ax.  2.) 


Then  7= 

8  4 

Multiply  by  8,  and 

3X—  .11  =10a>—  10+440—  56x  ; 
Transposing  and  uniting  terms,  we  have 

49#=441  ; 
By  division,  x=9, 

6.  Given    §ar+2^  +  ll=fa?-f  17,    to  find   x. 

If  we  commence  by  clearing  of  fractions,  we  should  make 
comparatively  a  long  and  tedious  operation.  Let  us  first  reduce 
it  by  striking  out  equals  from  both  sides  of  the  equation.  We 
can  take  1  1  from  both  sides  without  any  formality  of  transposing 
or  changing  signs  ;  say  drop  equals  from  both  sides,  (Ax.  2.)  and 
reduce  the  fraction  f#=|a?. 

All  this  can  be  done  as  quick  as  thought,  and  we  shall  have 

Multiply  by  4,  then 


+10=3+24,    or          =a 
5  o 

Hence,  7#=70,         or         x—lQ  Am. 

7.  Given    %x—  5-KaM-8-Kx—  10=100—  6—  7   to  f.nd  the 
value  of  x. 

Collecting  and  uniting  the  numeral  quantities,  we  have 


-^=; 
Multiply  every  term  by  60,  and  we  have 


Collecting  terms,  47a?=94.60 

Divide  both  sides  by  47,  and      x=  2.60=120  Ans. 


58  ELEMENTS  OF  ALGEBRA. 

(Art.  39.)  When  equations  contain  compound  fractions  and 
simple  ones,  clear  them  of  the  simple  fractions  first,  and  unite, 
as  far  as  possible,  all  the  simple  terms. 


EXAMPLES. 


6#-f-7  ,   7x—  13 

8.  Given  ~—  ~-+  a    ,  Q  =•—  ~-  to  find  the  value  of  a\ 
y        tx£-po        o 


Multiply  all  the  terms  by  the  smallest  denominator,  3.     That 
is,  divide  all  the  denominators  by  3,  and 


Multiply  by  3  again,  and  6a?+7+  0  " 

Drop  6#-f-7,  and  -  -         =5. 
2>x~\~  1 

Clear  of  fractions,   21  a:  —  39=10a?-f  5. 

Drop  IQx  and  add  39,  and  we  have  11#=44,  or  x=4. 

9.  Given   —  —  --  -  --  =-  to  find  the  value  of  x. 
21         4x  —  11     3 

Observe  that  —  ~  —  may  be  expressed  in  two  parts,  thus, 

7x     16  7x     x 

—  -j-ry     Observe  also,  that  —  -=-.     Hence  these  terms  may 
2121  2  1      o 

be  dropped,  the  remainders    must  be   equal.     Transpose    the 

16       x-\-S 

minus  term,  then  —  =  --  . 
21     4x  —  11 

Clear  of  fractions,  and  64^—11X16=21^4-21X8. 

Drop  21a;  and  observe  that  11  X  16  is  the  same  as  22  X  8. 

Then     432—-  22X8=21X8.         Let  «=8, 

Then     43z—  22a=21«. 

Transpose  —  22a   and   43#=43a. 

Hence  x=a.     But  a=8.     Therefore  x=8. 

N.  B.  We  operate  thus,  to  call  attention  to  the  relation  of 
quantities,  and  to  form  a  habit  of  quick  comparison,  which  will, 
in  many  instances,  save  much  labor  and  introduce  the  pupil  into 
the  true  spirit  of  the  science 


EQUATIONS.  59 

1O.  Given = r~\"7  to  find  tne  value  of  or. 

36          ox — 4      4 

9a? 

By  a  slight  examination  we  perceive  that  —  is   equal   to   %x. 

oD 

Hence  these  terms  may  be  left  out,  as  they  balance  each  other. 


Therefore 


Clear  of  fractions,  and   25a?  —  20  =36#  —  108. 
Transpose  25x  —  36#=20  —  108. 

Unite  and  change  signs,  and    11#=88  or  a?=8,     Jim 


.  36  .  5#-}-20     4x  .  86  t 
11.  Glven  _+_+§_is=T+_  to  find  *. 

By  taking  equals  from  both  sides,  we  have 

5#t-20 

-  -  rz—%'     By  reduction  a?  =4. 
Qx  —  16 


12.  Given  -  ---  ^—=6^  --  r  —  to  find  *• 

42  4 

Multiply  by  4,  to  clear  of  fractions,  and 

3tf  —  2#-f-2=24;r—  20#  —  13.     Reduced   x=5. 

(Art.  40.)  When  a  minus  sign  stands  before  a  compound 
quantity,  it  indicates  that  the  whole  is  to  be  subtracted  ;  but  we 
subtract  by  changing  signs,  (Art.  5).  The  minus  sign  before 

y.  __  I 

-   in  the  last  example,  does  not  indicate  that  the  x  is  minus, 

but  that  this  term  must  be  subtracted.  When  the  term  is  multi- 
plied by  4,  the  numerator  becomes  2x  —  2,  and  subtracting  it 
we  have  —  2#-|-2. 

Having  thus  far  explained,  we  give  the  following  unwrought 
equations,  for  practice  : 

O/v»  n\ 

13.  Given  —=-+24  to  find  the  value  of  x.        Jns.  19],. 

14.  Given  zx-\-lx=lQ  to  find  the  value  of  x.       rfns.  24. 


60  ELEMENTS  OF  ALGEBRA. 

15.  Given  ^-+|.=20—^i_  to  find  z.  Ana     9 

16.  Given  ^±i+^±?=16-?  to  find  *.        Am.  13 


.  Given  2*-+,5=  to  find  *. 

3  5 


,a    ~.  2#+l     a?+3 

18.  Given  x  ---  _=__   to  find  x.         tins.    x=l3 


19.  Given  5#+5#+*tf+ya?=77  to  find  *•  ^s.  #=60. 

20.  Given  5a?+|a>Ha?=130  to  find  a?.  rfns.  x=120 

21.  Given  ^af+^+T^a^QO   to  find  x.  Jlns.  x=l20 

22.  Given  ^+^+^=82   to  find  y.  ^ns.  y=84 

23.  Given  5a?-FIa;+5a?=34   to  find  x.  tins. 


24.  Given    11^4-+++=315   to  find  x. 


25.  Given  3/++++=146  to  find  y.   Ans.  y=56 


26.  Given  _29=_-30  to  find  a?,     ^n*.  a?=2 

ar+2  a?  —  2 

There  is  a  peculiar  circumstance  attending  this  26th  example, 
and  the  4th  example  of  Art.  42,  which  will  cause  us  to  refer  to 
them  in  a  subsequent  part  of  this  work. 

N.  B.  In  solving  equations  19,  20,  21,  22,  and  23,  use  no 
larger  numbers  than  those  given,  indicating  and  not  performing 
numeral  multiplications. 

(Art.  41.)  Every  proportion  may  be  converted  into  an  equa- 
tion. Proportion  is  nothing  more  than  an  assumption  that  the 
same  relation  or  the  same  ratio  exists  between  two  quantities, 
as  exists  between  two  other  quantities. 

That  is,  Ji  is  to  B  as  C  is  to  D.  There  is  some  relation  be- 
tween A  and  B.  Let  r  express  that  relation,  then  /?=r*#.  But 


EQUATIONS.  61 

the  relation  between  C  and  D  is  the  same  (by  hypothesis)  as 
between  Ji  and  B.     Hence  D=rC. 

Then  in  place  of  A  :  B  :  :  C:D 
we  have    Ji  \rA\\  C:rC 


Multiply  the  extreme  terms,  and  we  have 

Multiply  the  mean  terms,  and  we  have 

Obviously  the  same  product,  whatever  quantities  may  be  re- 
presented by  either  .#,  or  r,  or  C. 

Hence,  to  convert  a  proportion  into  an  equation,  we  have  the 
following 

RULE.  Place,  the  product  of  the  extremes  equal  to  the  pro- 
duct  of  the  means. 

(Art.  42.)  The  relation  between  two  quantities  is  not  changed 
by  multiplying  or  dividing  both  of  them  by  the  same  quantity. 
Thus,  a  :  b  ::  2a  :  2b,  or  more  generally,  a  :  b  ::  na  :  nb,  for  the 
product  of  the  extremes  is  obviously  equal  to  the  product  of  the 
means. 

That  is,  a  is  to  b  as  any  number  of  times  a  is  to  the  same 
number  of  times  b. 

We  shall  take  up  proportion  again,  but  Articles  41  and  42  are 
sufficient  for  our  present  purpose. 

EXAMPLES. 

1.  Given   3.r—  1  :  2;r-f-l  ::  3x  :  x   to  find   x. 
(By  Art.  41.)         3x2—  x=6x*+3x. 
Transpose  and  unite,  and  we  have   Q=3x*-{-4x. 
Divide  by  x,  and     3x4-4=0     or     x—  —  f  ,     Jlns. 

3      3x 

2.  Given   '-  :  —  :  :  6  :  5x  —  4   to  find   x. 

The  first  two  terms  have  the  same  relation  as  5  :  Jar,  or  oa 
2  :  x.  Hence  2  :  x  ::  6  :  5x  —  4. 

Product  of  extremes  and  means,    lOa:  —  S=Gx   or   x=Z. 


X. 

Am.    #=:3. 


02  ELEMENTS  OF  ALGEBRA. 

y          K 

4.  Given   —  -  :  (#—5) : :  f  :  £   to  find   x. 

•fins.     £-=5, 

5.  Given  a?-{-2  :  a  : :  b  :  c  to  find  the  value  of  x. 

dns.     j?= 2. 

c 

6.  Given  2a? — 3  :  x — 1  ::  2x  :  #-f-l    to  find  the  value  of  x. 

Jlns.     x=3 

7.  Given   z-J-6  :  38 — x  : :  9  :  2   to  find   x.          fins.  x=30. 

8.  Given   #+4  :  x — 11  : :  100  :  40    to  find   x.     Jlns.  x=2l. 

QUESTIONS    PRODUCING    SIMPLE    EQUATIONS. 

(Art.  43.)  We  now  suppose  the  pupil  can  readily  reduce  a 
simple  equation  containing  but  one  unknown  quantity,  and  he  is, 
therefore,  prepared  to  solve  the  following  questions.  The  only 
difficulty  he  can  experience  is  the  want  of  tact  to  reason  briefly 
and  powerfully  with  algebraic  symbols ;  but  this  tact  can  only 
be  acquired  by  practice  and  strict  attention  to  the  solution  of 
questions.  We  can  only  give  the  following  general  direction : 

Represent  the  unknown  quantify  by  some  symbol  or  letter, 
and  really  consider  it  as  definite  and  known,  and  go  over  the 
same  operations  as  to  verify  the  answer  when  known. 

EXAMPLES. 

1.  What  number  is  that  whose  third  part  added  to  its  fourth 
part  makes  21 1  Jlns.     36. 

The  number  may  be  represented  by  x. 

Then   $x-}-Zx=2l.  Therefore  x=3G. 

2.  Two  men  having  found  a  bag  of  money,  disputed  about 
the  division  of  it.     One  said  that  the  half,  the  third,  and  the 
fourth  parts  of  it  made  $130,  and  if  the  other  could  tell  how 
much  money  the  bag  contained,  he  might  have  it  all.     How 
much  money  did  the  bag  contain  1  Jlns.  $120. 

(See  equation  20,  Art.  40.) 


EQUATIONS.  63 

3.  A  man  has  a  lease  for  20  years,  one-third  of  the  time  pas. 
is  equal  to  one-half  of  the  time  to  come.     How  much  of  the  time 
nas  passed  ? 

Let  x=  the  time  past. 

Then  20 — x=  the  time  to  come. 

x     20—3? 

By  the  question  --= —    — .     Therefore   x=l2    Am. 
o          Z 

4.  What  number  is  that,  from  which  6  being  subtracted,  and 
the  remainder  multiplied  by  11,  the  product  will  be  121? 

Let  0:=  the  number. 

Then  (a: — 6)11=121,     or     x — 6=11    by  division. 

Hence  x=l7. 

5.  It  is  required  to  find  two  numbers,  whose  difference  is  6, 
and  if  |  of  the  less  be  added  to  £  of  the  greater,  the  sum  will 
be  equal  to  ^  of  the  greater  diminished  by  £  of  the  less. 

Let  x—  the  less.     Then  #-f-6=  the  greater. 

By  the  question  %x-\ — — =— ±x. 

5  o 

Drop  |a?  from  both  sides  and  add  \x  to  both  sides,  and  we 

have   — £~-   =2,     or    x=2,   the  less  number. 
0 

We  may  clear  of  fractions  in  full,  and  then  transpose  and  unite 
terms,  but  the  operation  would  be  much  longer. 

6.  After  paying  |  and  ^  of  my  money,  I  had  $66  left ;  how 
much  had  I  at  first?  JLns.     $120. 

7.  After  paying  away  5  of  my  money,  and  |  of  what  remained, 
and  losing  ^  of  what  was  left,  I  found  that  I  had  still  $24.     How 
much  had  I  at  first  ?  rfns.    60. 

8.  What  number  is  that  from  which  if  5  be  subtracted,  §  of 
the  remainder  will  be  40  ?  Jlns.     65, 

9.  A  man  sold  a  horse  and  a  chaise  for  $200 ;  5  of  the  price 
of  the  horse  was  equal  to  |  of  the  price  of  the  chaise.     What 
was  the  price  of  each?  Jlns.  Chaise  $120.     Horse  $80. 


(M  ELEMENTS  OF  ALGEBRA. 

1O  Divide  48  into  two  such  parts,  that  if  the  less  be  divided 
by  4,  and  the  greater  by  6,  the  sum  of  the  quotients  will  be  9. 

rfns.     12  and  36. 

11.  An  estate  is  to  be  divided  among  4  children,  in  the  fol- 
lowing manner : 

The  first  is  to  have  $200  more  than  |  of  the  whole. 
The  second  is  to  have  $340  more  than  ^  of  the  whole. 
The  third  is  to  have  $300  more  than  i-  of  the  whole. 
And  the  fourth  is  to  have  $400  more  than  j  of  the  whole. 
What  is  the  value  of  the  estate  ?  Ans.     $4800 

12.  Find  two  numbers  in  the  proportion  of  3  to  4,  whose 
sum  shall  be  to  the  sum  of  their  squares  as  7  to  50. 

Jlns.     6  and  8. 

N.  B.  When  proportional  numbers  are  required,  it  is  generally 
most  convenient  to  represent  them  by  one  unknown  term,  with 
coefficients  of  the  given  relation.  Thus,  numbers  in  proportion 
of  3  to  4,  may  be  expressed  by  3x  and  4x,  and  the  proportion 
of  a  to  b  may  be  expressed  by  ax  and  bx. 

13.  The  sum  of  $2000  was  bequeathed  to  two  persons,  so 
that  the  share  of  A  should  be  to  that  of  B  as  7  to  9.     What  was 
the  share  of  each?          Jins.  «#'s  share  $875,  B's  share  $1125. 

14.  A  certain  sum  of  money  was  put  at  simple  interest,  and 
in  8  months  it  amounted  to  $1488,  and  in  15  months  it  amounted 
to  $1530.     What  was  the  sum  1  Ans.     $1440. 

Let  a?=  the  sum.  The  sum  or  principal  subtracted  from  the 
amount  will  give  interest:  therefore  1488 — x  represents  the 
interest  for  8  months,  and  1530 — x  is  the  interest  for  15  months. 

Now  whatever  be  the  rate  per  cent,  double  time  will  give 
double  interest,  &c.  Hence  8  :  15  ::  1488 — x  :  1530— x. 

N.  B.  To  acquire  true  delicacy  in  algebraical  operations,  it  is 
often  expedient  not  to  use  large  numerals,  but  let  them  be  repre- 
sented by  letters.  In  the  present  example  let  «=1488  Then 
a-f-42=1530,  and  the  proportion  becomes  8:  15::  a — z:a-\- 
42 — x. 


EQUATIONS.  05 

Multiply  extremes,  &c.,   8«-f-8-42 — Sx=l5a—I5x. 

Drop  Sa  and  — Sx.     We  then  have   8'42=7a — 7#. 

Dividing  by  7  and  transposing  x—a — 48=1440,     Am. 

15.  A  merchant  allows  $1000  per  annum  for  the  expenses 
of  his  family,  and  annually  increases  that  part  of  his  capital 
which  is  not  so  expended  by  a  third  of  it ;  at  the  end  of  three 
years  his  original  stock  will  double.  What  had  he  at  first  / 

rfns.     $14,800. 

Let  x=  the  original  stock,  and  a=1000. 

To  increase  any  quantity  by  its  £  part  is  equivalent  to  multi- 

\  3C       \Ct 

plying  it  by  £.     Hence  — - —  is  his  2d  year's  stock. 


16.  A  man  has  a  lease  for  99  years,  and  being  asked  how 
much  of  it  had  already  expired,  answered  that  f  of  the  time  past 
was  equal  to  £  of  the  time  to  come.     Required  the  time  past  and 
the  time  to  come.      • 

Assume  #=99.     Am.  Time  past,  54  years. 

17.  In  the  composition  of  a  quantity  of  gunpowder 
The  nitre  was  lOlbs.  more  than  f  of  the  whole, 


~r     . 
The  sulphur  4£  Ibs.  less  than  \  of  the  whole, 

The  charcoal  2  Ibs.  less  than  \  of  the  nitre. 

What  was  the  amount  of  gunpowder?  Jlns.     69 Ibs. 

18.  Divide  $183  between  two  men,  so  that  y  of  what  the  first 
receives  shall  be  equal  to  T\  of  what  the  second  receives.     What 
will  be  the  share  of  each  ?  Ans.     1st,  $63 ;  2d,  $120. 

19.  Divide  the  number  68  into  two  such  parts  that  the  differ- 
ence between  the  greater  and  84  shall  be  equal  to  3  times  the 
difference  between  the  less  and  40.     Ans.  Greater,  42  ;  Less,  26. 

20.  Four  places  are  situated  in  the  order  of  the  letters  A,  B, 
C,  D      The  distance  from  A  to  D  is  34  miles.     The  distance 
from  A  to  B  is  to  the  distance  from  C  to  D  as  2  to  3.     And  4 
of  the  distance  from  A  to  J?,  added  to  half  the  distance  from  C 
to  D,  is  three  times  the  distance  from  B  to  C.     What  are  the 
respective  distances  ? 

Ans.  From  A  to  7?=12 ;  from  B  to  C=4  ;  frorp  C  to//=18, 
6 


66  ELEMENTS  OF  ALGEBRA. 

21.  A  man  driving  a  flock  of  sheep  to  market,  was  met  by  a 
party  of  soldiers,  who  plundered  him  of  5  of  his  flock  and  0 
more.     Afterwards  he  was  met  by  another  company,  who  took 
5  what  he  then  had  and  10  more:  after  that  he  had  but  2  left. 
How  many  had  he  at  first?  fins.     45. 

22.  A  laborer  engaged  to  serve  for  60  days  on  these  comli 
tions :  That  for  every  day  he  worked  he  should  have  75  cents 
and  his  board,  and  for  every  day  he  was  idle  he  should  forfeit  25 
cents  for  damage  and  board.     At  the  end  of  the  time  a  settlement 
was  made  and  he  received  $25.     How  many  days  did  he  work, 
and  how  many  days  was  he  idle  ? 

The  common  way  of  solving  such  questions  is  to  let  x=  the 
days  he  worked ;  then  60 — x  represents  the  days  he  was  idle. 
Then  sum  up  the  account  and  put  it  equal  to  $25. 

Another  method  is  to  consider  that  if  he  worked  the  whole  60 
days,  at  75  cents  per  day,  he  must  receive  $45.  But  for  every 
day  he  was  idle,  he  not  only  lost  his  w^es,  75  cents,  but  25 
cents  in  addition.  That  is,  he  lost  $1  every  day  he  was  idle. 

Now  let  x=  the  days  he  was  idle.  Then  x=  the  dollars 
he  lost.  And  45 — #=25  or  #=20  the  days  he  was  idle. 

23.  A  boy  engaged  to  carry  100  glass  vessels  to  a  certain 
place,  and  to  receive  3  cents  for  every  one  he  delivered,  and  to 
forfeit  9  cents  for  every  one  he  broke.     On  settlement,  he  re- 
ceived 2  dollars  and  40  cents.     How  many  did  he  break  ? 

rfns.     5. 

24.  A  person  engaged  to  work  a  days  on  these  conditions : 
For  each  day  he  worked  he  was  to  receive  b  cents,  for  each  day 
he  was  idle  he  was  to  forfeit  c  cents.     At  the  end  of  a  days  he 
received  d  cents.     How  many  days  was  he  idle  ? 

ab— d  . 

Jlns.    -T-, —  days, 
o-f-c 

25.  It  is  required  to  divide  the  number  204  into  two  such 
parts,  that  f  of  the  less  being  taken  from  the  greater,  the  remain- 
der will  be  equal  to  f  of  the  greater  subtracted  from  4  times  the 
less.  Arts.     The  numbers  are  154  and  50.* 


EQUATIONS.  67 

(Art.  44.)  We  introduce  this,  and  a  few  following  problems, 
to  teach  one  important  expedient,  not  to  say  principle,  which  is, 
not  always  to  commence  a  problem  by  putting  the  unknown 
quantity  equal  to  a  single  letter.  We  may  take  2;r,  3;r,  or  nx 
to  represent  the  unknown  quantity,  as  well  as  #,  and  we  may 
resort  to  this  expedient  when  fractional  parts  of  the  quantity  are 
called  in  question,  and  take  such  a  number  of  ar's  as  may  be 
divided  without  fractions. 

In  the  present  example  we  do  not  put  x=  to  the  less  part,  as 
we  must  have  f  of  the  less  part.  It  will  be  more  convenient  to 
put  5x=  the  less  part.  Then  f  of  it  will  be  2x.  Put  a=204. 

26.  A  man  bought  a  horse  and  chaise  for  341  («)  dollars. 
Now  if  J  of  the  price  of  the  horse  be  subtracted  from  twice  the 
price  of  the  chaise,  the  remainder  will  be  the  same  as  if  |  of  the 
price  of  the  chaise  be  subtracted  from  3  times  the  price  of  the 
horse.     Required  the  price  of  each. 

Ans.     Horse  $152.    Chaise  $189. 

N.  B.  Let  Sx=  the  price  of  the  horse. 
Or  let  7x=  the  price  of  the  chaise. 
Solve  this  question  by  both  of  these  notations. 

27.  From  two  casks  of  equal  size  are  drawn  quantities,  which 
are  in  the  proportion  of  6  to  7;  and  it  appears  that  if  16  gallons  less 
had  been  drawn  from  that  which  now  contains  the  less,  only  one 
half  as  much  would  have  been  drawn  from  it  as  from  the  other. 
How  many  gallons  were  drawn  from  each  1        Jlns.  24  and  28. 

N.  B.  Let  Qx  and  7x  equal  the  quantities  drawn  out. 

28.  Divide  $315  among  four  persons,  ./?,  B,  C,  and  D,  giving 
B  as  much  and  £  more  than  Jl ;  C  5  more  than  Jl  and  B  toge- 
ther ;  and  D  |  more  than  A,  B  and  C.     What  is  the  share  of 
eadi  ?  Ans.     ./?  $24.     B  $36.     C  $80,  and  D  $175. 

If  we  take  x  to  represent  «#'s  share,  we  shall  have  a  very 
complex  and  troublesome  problem.*  But  it  will  be  more  simple 
by  making  6x=J2's  share. 

*  Taking  x  for  yTs  share,  and  reducing  their  sum,  gives  Equation  24, 
Art.  40. 


68  ELEMENTS  OF  ALGEBRA. 

Thus,  let  6x=tfs  share. 

Then  9x—JB%s  share. 

And  l5z-\-5x=C's  share. 


Also  35#-j—  —  =Z)'s  share. 


Sum  70*4—^=315 

280oH-35#=315X4 
315^=315X4 

x=4     Hence   6a?=24,  *#'s  sh. 

549.  A  gamester  at  play  staked  ^  of  his  money,  which  he  lost, 
but  afterwards  won  4  shillings  ;  he  then  lost  ?  of  what  he  had, 
and  afterwards  won  3  shillings  ;  after  this  he  lost  ^  of  what  he 
had,  and  finding  that  he  had  but  20  shillings  remaining,  he  left 
off  playing.  How  much  had  he  at  first  ?  rfns.  30  shillings. 

30.  A  gentleman  spends  §  of  his  yearly  income  for  the  sup- 
port of  his  family,  and  J  of  the  remainder  for  improving  his 
house  and  grounds,  and  lays  by  $70  a  year.     What  is  his  in- 
come?    Jlns.   9X70    dollars,  or  more  generally,  9  times  the 
sum  he  saves. 

31.  Divide  the  number  60  (a)  into  two  such  parts  that  their 
product  may  be  equal  to  three  times  the  square  of  the  less  num- 
ber ?  Ans.  15  and  45,  or  |a=  the  less  part. 

32.  After  paying  away  ?  and  \  of  my  money,  I  had  34  (a) 
dollars  left.     What  had  I  at  first  ? 

rfns.  56  dollars.     General  answer  y^X28. 

33.  My  horse  and  saddle  are  together  worth  90  (a)  dollars, 
and  my  horse  is  worth  8  times  my  saddle.     What  is  the  value 
of  each  ?  rfns.     Saddle  $10.     Horse  $80. 

34.  My  horse  and  saddle  are  together  worth  a  dollars,  and 
my  horse  is  worth  n  times  my  saddle.     What  is  the  value  of 

each  ?  Jlns.     Saddle      r        Horse 


EQUATIONS.  69 

35.  The  rent  of  an  estate  is  8  per  cent  greater  this  year  than 
last.     This  year  it  is  1890  dollars.     What  was  it  last  year? 

Ans.  $1750. 

36    The  rent  of  an  estate  is  n  per  cent,  greater  this  year  than 
last.     This  year  it  is  a  dollars.     What  was  it  last  year  1 

100« 

dollare' 

37.  A  and  B  have  the  same  income.     A  contracts  an  annual 
debt  amounting  to  1  of  it  ;  B  lives  upon  |  of  it  ;  at  the  end  of 
two  years  B  lends  to  A  enough  to  pay  off  his  debts,  and  has  32 
(a)  dollars  to  spare.     What  is  the  income  of  each  ? 

Ans.     $280  or  £(35a). 

38.  What  number  is  that  of  which  |,  |  and  f  added  together 

make  73  (a)  ?  .   a       84a 

v  '  Am.  84.     General  Ans.  ——. 

73 

39.  A  person  after  spending  100  dollars  more  than  I  of  his 
income,  had  remaining  35  dollars  more  than  5  of  it.     Required 
his  income.  Ans.  $450. 

40.  A  person  after  spending  (a)  dollars  more  than  \  of  his 
income,  had  remaining  (6)  dollars  more  than  £  of  it.     Required 
his  income. 


41.  There  are  two  numbers  in  proportion  of  2  to  3,  and  if  4 
be  added  to  each  of  them,  the  sums  will  be  in  proportion  of  5 
to  7?  Ans.  16  and  24. 

42.  It  is  required  to  find  a  number  such,  that  if  it  be  increased 
by  7,  the  square  root  of  the  sum  shall  be  equal  to  the  square 
root  of  the  number  itself,  and  1  more.  Ans.     9. 

43.  A  sets  out  from  a  certain  place,  and  travels  at  the  rate  of 
7  miles  in  5  hours  ;  and  8  hours  afterward  B  sets  out  from  the 
same  place  in  pursuit,  at  the  rate  of  5  miles  in  3  hours.     How 
long  and  how  far  must  B  travel  before  he  overtakes  A  ? 

Ans.     42  hours,  and  70  raijes 


70  ELEMENTS  OF  AUJEURA. 

SIMPLE  EQUATIONS. 
CHAPTER  II. 

(Art.  45.)  We  have  given  a  sufficient  number  of  examplost 
and  introduced  the  reader  sufficiently  far  into  the  science  pre- 
vious to  giving  instructions  for  the  solution  of  questions  contain- 
ing two  or  more  unknown  quantities. 

There  are  many  simple  problems  which  one  may  meet  with 
in  algebra  which  cannot  be  solved  by  the  use  of  a  single  un- 
known quantity,  and  there  are  also  some  which  may  be  sohid 
by  a  single  unknown  letter,  that  may  become  much  more  simple 
by  using  two  or  more  unknown  quantities. 

When  two  unknown  quantities  are  used,  two  independent 
equations  must  exist,  in  which  the  value  of  the  unknown  letters 
must  be  the  same  in  each.  When  three  unknown  quantities  are 
used,  there  must  exist  three  independent  equations,  in  which  the 
value  of  any  one  of  the  unknown  letters  is  the  same  in  each. 

In  short,  there  must  be  as  many  independent  equations  as 
unknown  quantities  used  in  the  question. 

For  more  definite  illustration  let  us  suppose  the  following 
question : 

ji  merchant  sends  me  a  bill  of  16  dollars  for  3  pair  of  shoes 
and  2  pair  of  boots  ;  afterwards  he  sends  another  bill  of  23 
dollars  for  4  pair  of  shoes  and  3  pair  of  boots,  charging  at  the 
same  rate.  Wliat  was  his  price  for  a  pair  of  shoes,  and  what 
for  a  pair  of  boots  ? 

This  can  be  resolved  by  one  unknown  quantity,  but  it  is  far 
more  simple  to  use  two.  , 

Let  x=  the  price  of  a  pair  of  shoes, 

And  y=  the  price  of  a  pair  of  boots. 

Then  by  the  question  3a?-|-2y=16 

And  4*-}-3r/=23. 

These  two  equations  are  independent ;  that  is,  one  cannot  be 
converted  into  the  other  by  multiplication  or  division,  notwith- 
standing the  value  of  a:  and  of  y  is  the  same  in  both  equations. 

Having  intimated  that  this  problem  can  be  resolved  with  one 


EQUATIONS.  7! 

unknown  quantity,  we  will  explain  in  what  manner,  before  we 
proceed  to  a  general  solution  of  equations  containing  two  un- 
known quantities. 

Let  #=  the  price  of  a  pair  of  shoes. 
Then  3x=  the  price  of  three  pair  of  shoes. 
And  16—  3x—  the  price  of  two  pair  of  boots. 

•I    /»  O/y» 

Consequently  -  =  the  price  of  one  pair  of  boots. 
Now  4  pair  of  shoes  which  cost  4#,  and  3  pair  of  boots  which 
cost  -  being  added  together,  must  equal  23  dollars. 

That  is,     4x4-24—^=23. 

Or,  1  —  £#=0.     Therefore  a?—  2  dollars,  the  price 

of  a  pair  of  shoes.     Substitute  the  value  of  x  in  the  expression 

1  A        Q'v* 

-  and  we  find  5  dollars  for  the  price  of  a  pair  of  boots. 
2 

Now  let  us  resume  the  equations, 


=23     (B) 


FIRST    METHOD    OF    ELIMINATION. 

(Art.  46.)  Transpose  the  terms  containing  y  to  the  right  hand 
sides  of  the  equations,  and  divide  by  the  coefficients  of  x,  and 

From  equation  (rf)  we  have  x= — — - -  (C) 

o 

And  from  (B)  we  have  x=- — ~^-^-  (B) 

4 

Put  the  two  expressions  for  x  equal  to  each  other.     (Ax.  7.) 
And  — — —  ==  — - — . 

An  equation  which  readily  gives  y=5,  which,  taken  as  the 
value  of  y,  in  either  equation  (C)  or  (Z>)  will  give   #=2. 

This  method  of  elimination,  just  explained,  is   called  th  > 
method  by  comparison. 


72  ELEMENTS   OF  ALGEBRA. 

SECOND    METHOD    OF    ELIMINATION. 
(Art.  47.)  To  explain  another  method  of  solution,  let  us  again 
resume  the  equations : 

3x-\-2y=lG        (A) 
4#-f3i/=23         (if) 

The  value  of  x  from  equation  (»#)  is  x=i(lQ — 2y). 

Substitute  this  value  for  x  in  equation  (2?),  and  we  have 
4X  £(16 — 2y)+3y^=23,  an  equation  containing  only  y. 

Reducing  it,  we  find  y=5  the  same  as  before. 

This  method  of  elimination  is  called  the  method  by  substitu- 
tion, and  consists  in  finding  the  value  of  one  unknown  quantity 
from  one  equation  to  put  that  value  in  the  other  which  will  cause 
one  unknown  quantity  to  disappear. 

THIRD    METHOD    OF    ELIMINATION. 

(Art.  48.)  Resume  again  Sx-\-2y=l6         (.#) 
4x+3y=23         (£) 

When  the  coefficients  of  either  x  or  y  are  the  same  in  both 
equations,  and  the  signs  alike,  that  term  will  disappear  by  sub- 
traction. 

When  the  signs  are  unlike,  and  the  coefficients  equal,  the  term 
will  disappear  by  addition. 

To  make  the  coefficients  of  x  equal,  multiply  each  equation 
by  the  coefficient  of  x  in  the  other. 

To  make  the  coefficients  of  y  equal,  multiply  each  equation 
ly  the  coefficient  of  y  in  the  other. 

Multiply  equation  (A)  by  4  and  l2x-\-8y=Q4 
Multiply  equation  (E)  by  3  and  12#-|-9i/=69 

Difference  y  =  5  as  before. 

To  continue  this  investigation,  let  us  take  the  equations 
2x+3y=23         (A) 
5x— 2y=iO         (JB) 

Multiply  equation  (A)  by  2,  and  equation  (B)  by  3,  and  we 
have  4x-{-6y=46 


EQUATIONS,  73 


Equations  in  which  the  coefficients  of  y  are  equal,  and  the 
signs  unlike.  In  this  case  add,  and  the  y's  will  destroy  each 
other,  giving  19#-=76 

Or  a?=4. 

This  method  of  elimination  is  called  the  method  by  addition 
and  subtraction. 

FOURTH    METHOD    OF    ELIMINATION. 

(Art.  48.)  Take  the  equations     2x-\-3y=23.     (.#) 
And  5x— 2t/=10.     (£) 

Multiply  one  of  the  equations,  for  example  (,/#),  by  some  inde- 
terminate quantity,  say  m. 

Then  2mx-\-3my=23m 

Subtract  (£)  5x  —  2y=lO 

Remainder,  (C)  (2m — 5)a?+ (3m+2)y=23m— 10 

As  m  is  an  indeterminate  quantity,  we  can  assume  it  of  any 
value  to  suit  our  pleasure,  and  whatever  the  assumption  may  be, 
the  equation  is  still  true. 

Let  us  assume  it  of  such  a  value  as  shall  make  the  coefficient 
of  y,  (3m-l-2)=0. 

The  whole  term  will  then  be  0  times  ?/,  which  is  0,  and  equa- 
tion (C)  becomes 

(2m  -5)#=23m — 10 

23m — 10         ,  nx 

°f  *=-2^f          W 

But  3w-f2=0.         Therefore  «i= — |. 
Which  substitute  for  m  in  equation  (/)),  and  we  have 

__23  X  f — 10  _— 23  X  2^30_— 76^i 
x~  —2Xf— -5   =~  —  2X2— 15~— -19~~ 

This  is  a  French  method,  introduced  by  Bezout,  but  it  is  too 
indirect  and  metaphysical  to  be  much  practised,  or  in  fact  much 
known. 

Of  the  other  three  methods,  sometimes  one  is  preferable  and 
sometimes  another,  according  to  the  relation  of  the  coefficients 
and  the  positions  in  which  they  stand. 
7 


74  ELEMENTS  OF  ALGEBRA. 

No  one  should  be  prejudiced  against  either  method,  and  in 
practice  we  use  either  one,  or  modifications  of  them,  as  the  case 
may  require.  The  forms  may  be  disregarded  when  the  princi- 
ples are  kept  in  view. 

(Art.  49.)  To  present  these  different  forms  in  the  most  general 
manner,  let  us  take  the  following  general  equations,  as  all  par- 
ticular equations  can  be  reduced  to  these  forms. 

ax-\-by  =c        (J3) 

Observe  that  a  and  a',  may  represent  very  different  quantities, 
BO  b  and  b'  may  be  different,  also  c  and  c'  may  be  different.  In 
special  problems,  however,  a  may  be  equal  to  a',  or  be  some 
multiple  of  it ;  and  the  same  remark  may  apply  to  the  other 
letters.  In  such  cases  the  solution  of  the  equations  is  much 
easier  than  by  the  definite  forms.  Hence,  in  solving  definite 
problems  great  attention  should  be  paid  to  the  relative  values  of 
the  coefficients. 

First  method. 

Transpose  the  terms  containing  y  and  divide  by  the  coeffi- 
cients of  a?,  and 

also     x=c^y         (C) 
a' 

Therefore -=- —        (Axiom  7.) 

a  a 

Clearing  of  fractions,  give     a'c — a'by=acf — ab'y. 
Transpose,  and  (ab' — a'b)y=ac' — a'c. 

T»    j-  •  •  ac' — a'c 

By  division         y——r, 77- 

y     ab' — ab 

When  y  is  determined,  its  value  put  in  either  equation  marked 
(C)  will  give  x. 

Second  method. 
From  equation  (.#)  ' — ' 


EQUATIONS.  75 

Which  value  of  x  substitute  in  equation  (B)  and 
a'c  —  a'by 


Clearing  of  fractions  and  transposing  a'c,  we  have 

ab  'y  —  a'by=ac'  —  a'c 
~  ac'  —  a'c 

Ot  y=^=«b 

The  same  value  of  y  as  before  found. 

Third  method. 

Multiply  equation  (rf)  by  «',  and  equation  (B)  by  a. 
And  a'ax-\-a'by=a'c. 

Also  a'ax-}-ab'y=ac' 

Difference  (ab'  ~a'b}y=ac'  —  a'c 

_  ac'  —  a'c  . 

Or  y=—r,  --  TL  same  va^ue  as  by  the 

ab'  —  ab 

Dther  methods. 

Fourth  method. 

Multiply  equation  (JT)  by  an  indefinite  number  m, 
And  amx-}-bmy=mc 

Subtract  (B]  a'x-\-   b'y=c' 


And  (am  —  a']x-}-(bm  —  b']y=mc  —  c'. 

Now  the  value  of  m  may  be  so  assumed  as  to  render  the 
coefficient  of  #=0,     or     am  —  a'=0. 

Then  (bm  —  b')y=mc—c' 


But  am  —  a'=0,         or        m=—  . 

a 


76  ELEMENTS  OF  ALGEBRA. 

Put  this  resultant  value  of  m,  in  equation  ((7),  and 


cX 


ca'  —  ac' 


bX--b> 
a 

by  multiplying  both  numerator  and  denominator  by  a. 

(Art.  50.)  The  principles  just  explained  for  elimination  be- 
tween two  quantities  may  be  extended  to  any  number,  where  the 
number  of  independent  equations  given  are  equal  to  the  number 
of  unknown  quantities.  For  instance,  suppose  we  have  the 
three  independent  equations  : 

ax+by+cz=d  (A) 

a'x+b'y+c'z=d'          (B) 
a"x+b"y-}-c"z=d"      (C) 

We  can  eliminate  either  a?,  or  y,  or  z,  (whichever  may  be 
most  convenient  in  any  definite  problem)  between  equations  (A] 
and  (J5,)  and  we  shall  have  a  new  equation  containing  only  two 
unknown  quantities.  We  can  then  eliminate  the  same  letter  be- 
tween equations  (B}  and  ((7,)  or  (A]  and  (C*,)  and  have  another 
equation  containing  the  same  two  unknown  quantities. 

Then  we  shall  have  two  independent  equations,  containing  two 
unknown  quantities,  which  can  be  resolved  by  either  of  the  four 
methods  already  explained. 

(Art.  51.)  Another  theoretical  method,  from  the  French,  we 
present  to  the  reader,  more  for  curiosity  than  for  any  thing  else. 

Multiply  the  first  equation  (./?,)  by  an  indefinite  assumed  num- 
ber m.  Multiply  the  second  equation  (B]  by  another  indefinite 
number  n,  and  add  their  products  together.  Their  sum  will  be 

(am-}-a'n)x-\-(bm-\-b'n)y-i-(cm-{-c'n}z=dm-i-nd' 
Subtract  eq.  (C)      a"x  +b"y  c"z=d". 


And  (am-\-a'n  —  a"}x-\-(bm-\-b'n  —  &")7/-j-(cra4-c'n  —  c"}z 

=dm-\-nd'  —  d" 


EQUATIONS.  77 

As  m  and  n  are  independent  and  arbitrary  numbers,  they  can 
be  so  assumed  that 

am-\-a'n  —  «"=0         and        bm+b'n—  &"=0. 
Then          am-\-a'n=a"   (1)       and        bm-{-b'n=b"  (2) 


And  z=  -  :  --  -  (3) 

cm+c'n  —  c" 

From  equations  (1)  and  (2)  we  can  find  the  values  of  m  and 
»,  which  values  may  be  substituted  in  equation  (3,)  and  then  z 
will  be  fully  determined. 

EXAMPLES    FOR    PRACTICE. 

1.  Given  \          ,  >  to  find  the  values  of  x  and  y. 

?  12#+7=1005 


We  can  resolve  this  problem  by  either  one  of  the  four  methods 
just  explained.  But  we  would  not  restrict  the  pupil  to  the  very 
letter  of  the  rule,  for  that  in  many  cases  might  lead  to  operations 
unnecessarily  lengthy. 

If  we  take  the  third  method  of  elimination,  we  should  multi- 
ply the  first  equation  by  12,  the  second  by  8  ;  but  as  the  coeffi- 
cients of  x  contain  the  common  factor  4,  we  can  multiply  by  3 
and  2,  in  place  of  12  and  8.  That  is,  multiply  by  the  fourth 
part  of  12  and  8. 

In  practice  even  this  form  need  not  be  observed,  we  may  de- 
cide on  our  multipliers  by  inspection  only. 

Three  times  the  1st  gives  24#-f-15?/=204 

Twice  the  2d  gives  24a?-J-14i/=200 

Difference  gives  y—4 

Substituting  this  value  of  y  in  first  equation,  and 
8#-f-20=68         or         x=Q. 

In  solving  this,  we  have  used  modifications  of  the  3d  and  2d 
formal  methods. 


78  ELEMENTS  OF  ALGEBRA. 

For  exercise,  let  us  use  the  4th  method. 

8  mx  -|-  5my = 68m 
Take  l2x-\-  7y  =  100 


7)y=68m—  -100 
Assume        8m  —  12=0. 
68m—  100 

Then       =--        But  m==- 


„,,       P  68X1—100     204—200 

Therefore  =-__== 


2.  Given     \  ^J=  J9  J  to  find  *  and  y. 

If  we  multiply  the  first  of  these  equations  by  3,  the  coefficients 
of  y  will  be  equal,  and  the  equations  become 

15a?-r-6y=57, 
And  7a?  —  6?/=9. 

To  eliminate  y,  we  add  these  equations  (the  signs  of  the  terms 
containing  y  being  unlike),  and  there  results 


This  value  of  x  put  in  the  1st  equation  gives 


And  2/=2- 

3.  Given  ?-t^-J-6?/=21  and  01^4-5^=23  to  find  x  and  y 

Clear  of  fractions  and  reduce.     We  then  have 

#-{-24^=76 
And  15ff-f  y  =63. 

In  this  case  there  are  no  abbreviations  of  the  rules,  as  the 
coefficients  of  the  unknown  terms  are  prime  to  each  other. 
Continuing  the  operation,  we  find  a?=4,  y=3. 


EQUATIONS.  79 

2^?     3t/ 
4.  Given  #-{-?/=  17   and   —  ~~T  to  &n&  x  an(^  y« 

Owing  to  the  peculiarity  of  form  in  the  2d  equation,  it  is  most 
expedient  to  resolve  this  by  the  2d  method. 

From  the  2d,    x=^.         Then  ^-f  y=17. 

O  o 

Clearing  of  fractions,     9*/+8y  =  17  X  8. 

Or,  17y=17X8,     or    y=8. 

Hence,    a?=9. 


C  ^a:+8i/=1947 
5.  Given   <  J  —  isi  S  to  values  of  a?  and  y. 

Here  we  observe  that  both  x  and  y  are  divided  by  8,  x  in  one 
equation,  and  y  in  the  other  ;  also,  x  and  y  are  both  multiplied 
by  8. 

(Art.  51.)  All  such  circumstances  enable  us  to  resort  to  many 
pleasant  expedients  which  go  far  to  teach  the  true  spirit  of  al- 
gebra. 

Add  these  two  equations,  and  —  —  ^-f-8(#-}-y)=325. 

8 

Assume         x-\-y—s. 

Or  let  s  represent  the  sum  of  x-\-y,  then  js-f-8s=325. 
Clear  of  fractions,  and  s-{-  64s  =325X8. 
Unite  and  divide  by  65  and  5=5X8. 

Or  x-{-y=5a.  (A.)  By  returning  to  the  value  of  s,  and  put- 
ting a=8. 

Multiply  the  1st  equation  by  8,  and 


Subtract  (rf)  x  +  i/=5a 

Rem. 


Divide  by  63  and    y—3a=24.     Whence  #=2a=16. 
Let  the  pupil  take  any  one  of  the  formal  rules  for  the  solution 
of  the  preceding  equations,  and  mark  the  difference. 

6.  Given  $x-{-3y=2}  and    ^y-J-8.r=29  to  find  x  and  y. 

Jins.     o:=9.        =6 


80  ELEMENTS   OF  ALGEBRA. 

7.  Given  4#-f-t/=34  and  4y-\-x=l6,  to  find  ;raiid^. 

Ans.     x=S.    y=2 

8.  Given  5#-{-5?/=14  and  5a;+5y=ll,  to  find  x  and  y. 

?.     #=24.     v=6 


9.  Giver.  #-Hi/=8  and  zX-}-y=7  to  find  a1  and  y. 

Ans.     z=Q.    2/=4. 

10.  Given  \x+7y=99  and  iy-}-7,r=51  to  find  x  and  y. 

•flns.     x=7.       =14 


11.  Given 

4 

fins.     x=6Q.     i/=40 

12.  Given  -+-=6   and   —  |  —  =10,  to  find  x  and  y. 

x    y  x    y 

Multiply  the  first  equation  by  c,  the  second  by  a,  and  we  shall 
have 

etc  ,  be 
—]  ---  =6c 
x      y 

ac  .  ad 

--  ---  =10tf. 


By  subtraction     (be — ad)-—Qc — 10a 

f  be — ad 

Therefore =v. 

6c — 10a    y 


13.  Given  — — —=28  (Jft  and  — +— =—  (B)  to  find 
x        y  x      y      3    ^    ' 

tlie  values  of  x  and  y. 

21     21 

Divide  equation  (»#)  by  7,  and  — =4. 

x      y 

Divide  this  result  by  21,  and =--  (C\ 

x    y     21  ^    J 


EQUATIONS.  81 

Multiply  (C)  by  17,  gives  ^— H =5?  (/>) 

M>Q  210 

Subtract  (£)  from  (£)  and  we  have  — =— -. 

1      3      1 

Divide  by  73,  and  -=— =-     or    y=7.  • 

Putting  this  value  in  equation  (C)  and  reducing  we  find  x=3. 

14.  Given  -+-=- — 1    and   -+-=-+-  to  find  the  values 
x^y    y  x^y    0^2 

of  *  and  y  Jns.     *=4  and  y=2. 


?.     #=300.     7/=350. 

$3? — 24v2-H 
».  Given   3x+Gy+l=     0^     ^  ,  0 

to  find  a:  and  y. 

151— 1 6.r  ,  9^/—110 
And  3o?=— —  -\ —  H 


4y 1  3y 4 

.tfns.     a?=9.    y=2. 

In  the  first  equation  actually  divide  the  numerator  by  the  de- 
nominator, then  drop  equals  from  both  sides. 

17.  Given    ^    '  ~j!^"~L-oo  C  t0  find  the  values  of  x  and  ^* 

^ns.     a:=2.     i/=5 

18.  Given   <  ^  .  I  to  find  x  and  y. 

19.  Given  x-}-y =8  and  a?2—  y2=16  to  find  a:  and  t/. 

20.  Given  4(#-f-2/)=9(a;— y)  and  a?2— y2=36  to  find  *  and  y. 


82  ELEMENTS  OF  ALGEBRA. 

21.  Given  x  ly  ::  4  :  3   and   a?  —  ^2/3=37,   to  find  #  and  y. 

Ans.     #=4.    y=3. 

22.  Given  #-fr/=a  and  x*  —  yz=ab  to  find  x  and  y. 

0  a-\-b  a  —  /> 

Ans.    a?=—  .    2/=-2~ 

23.  Given  --  =£    and   -  -j—  =|  to  find  a;  and  y. 

Jlns.     #=4.       —lo 


24.  Given    $(a:+2)+  8y=31    and    4(y+5)-}-10^=192   to 
find  the  values  of  a;  and  y.  .tfns.     x=19.    y=3 

25.  Given  3a?+7y=79  and  2t/-j-5iC=19  to  find  the  values 
of  x  and  y.  Jlns.     #=10.    y=7. 

26.  Given    |(#-f  t/)+25=a?    and    %(x-}-y)  —  5=y   to    find 
the  values  of  x  and  y.  Jlns.     #=85.    y=35. 

27.  Given  x—  4=y+l  and  5#-^=^—  ^+37  to  find  the 

o       4        6 

values  of  x  and  y.  Ans.     x=S.    y—3 

28.  Given  4—  ^?=T/—  17|   and  ^=f+2     to    find     the 

D  O        O 

values  of  a?  and  y.  jJns.     x=W.    y=2Q. 

7x  —  21  .         x        .  3x  —  19  2a?+v 

29.  Given    —  —  -  }-y_=4-{—  ^—      and       -f-— 

9#—  7     3y4-9     4#+5y 
--  —  =-*-T  ---  rtr3  to  find  a?  and  y.       Ans.  #=9.  y=4. 


3O.  Given 


23—  #  2 


73—  3y 

---  2 


to  find  x  and 


3 
Jlns.     #=21. 


EQUATIONS.  83 

CHAPTER  HI. 

Solution  of  Equations  involving  three  or  more    unknown 
quantities. 

(Art.  52.)  No  additional  principles  are  requisite  to  those  given 
in  articles  49  and  50. 

EXAMPLES. 

f  X+    y+    z  =    9   ] 

1.  Given   \  #+2?/+3z—  16  [•  to  find  a?,  y,  and  z. 
[  a?+3y+4*=21  J 

By  the  1st  method,  transpose  the  terms  containing  y  and  z  in 
each  equation,  and 

x=  9—  T—  z, 


Then  putting  the  1st  and  2d  values  equal,  and  the  2d  and  3d 
values  equal,  gives 

9  —  y  —  z=16  —  2y  —  3z, 
1  6  —  2y—  3z  =2  1  —  3y—  40. 

Transposing  and  condensing  terms,  and 

y=7—2z, 

Also,  2/=5  —  z, 

Hence,       5  —  z=7  —  2z,         or        z=2 

Having  z=2,  we  have  2/=5  —  ^z=3,  and  having  the  values  of 
both  z  and  y,  by  the  first  equation  we  find  x=4. 

f  2x-\-4y—  3z=22  1 

a.  Given   <  4x—2y-}-5z=l8  [to  find  values  of  a?,  y  and  z 
[  Gx-}-7y—  z=63  J 

Multiplying  the  first  equation  by  2,         4x-}-Sy  —  6z~44 
And  subtracting  the  second,  4x  —  2y-\-  5z=18 

The  result  is,     (rf  l(—  llz=26 


54  ELEMENTS  OF  ALGEBRA. 

Then  multiply  the  first  equation  by  3,       Gx-{-l2y  —  9z=GG 
And  subtract  the  third,  6#-|-  7y  —    z  =63 

The  result  is,  (B)  5y—  Sz=  3 

Multiply  the  new  equation  (B)  by  2,  Wy  —  16z=  6 

And  subtract  this  from  equation  (.#)  IQy  —  llz=26 

The  result  is,  5z=20 

Therefore  z=  4 

Substituting  the  value  of  z  in  equation  (E]  and  we  find  y=7. 
Substituting  these  values  in  the  first  equation,  and  we  find  x=3. 

(    3x+9y-\-8z=4l  1 

3.  Given  \     5x-\-4y—  2z=2Q  \   to  find  a?,  y  and  z. 
[  llx+ty—  6^=37  J 

To  illustrate  by  a  practical  example  we  shall  resolve  this  by 
the  principles  explained  in  (Art.  51.) 

3mx-\-9my  -\-Smz  •=  41m 
5nx  -\-4ny  —  2nz  =2Qn 

Sum     (3ra+5n)#+(9m-f4n)?/-f-(8w  —  2n)z=41m+20w 
Take  llx  +7  —  6z=37 


Rem.    (3m4-5n—  11)  a;—  (7—  9m—  4n)y+(8m—  27i+6)z= 


Assume  3m-}-5n=ll         (1) 

And  9m-f4n=7  (2) 


From  equations  (1)  and  (2)  we  find  wi=  —  T3T  and  w=f  i 
These  values  substituted  in  equation  (3)  we  have 

_—  41  x  T3T4-2QX  f  f—  -37 
~  ~8XT3T—  2Xff+  6 

Multiply  both  numerator  and  denominator  by  11,  and  we  shall 

—123+520—407     —10 
have  z=- 


—24—  52-r-  66     —10 


EQUATIONS.  85 

Putting  this  value  of  z  in  the  1st  and  2d  equations,  we  shall 
have  only  two  equations  involving  x  and  y,  from  which  the 
values  of  these  letters  may  be  determined. 

These  equations  can  be  resolved  with  much  more  facility  by 
multiplying  the  2d  equation  by  4,  then  adding  it  to  the  1st  to 
destroy  the  terms  containing  z. 

Afterwards  multiplying  the  2d  equation  by  3,  and  subtracting 
the  3d  equation,  and  there  will  arise  two  equations  containing  x 
and  ?/,  which  may  be  resolved  by  one  of  the  methods  already 
explained. 

(Art.  53.)  When  three,  four,  or  more  unknown  quantities  with 
as  many  equations  are  given,  and  their  coefficients  are  all  prime 
to  each  other,  the  operation  is  necessarily  long.  But  when  sev- 
eral of  the  coefficients  are  multiples,  or  measures  of  each  other, 
or  unity,  several  expedients  may  be  resorted  to  for  the  purpose 
of  facilitating  calculation. 

No  specific  rules  can  be  given  for  mere  expedients.  Exam- 
ples alone  can  illustrate,  but  even  examples  will  be  fruitless  to 
one  who  neglects  general  principles  and  definite  theories.  Some 
few  expedients  will  be  illustrated  by  the  following 

EXAMPLES. 


1.  Given       x+y  —  z=25      to  find  x,  y,  and  z. 
[  x—y—z=  9  J 

Subtract  the  2d  from  the  1st,  and  2z=6. 
Subtract  the  3d  from  the  2d,  and  2?/=16. 
Add  the  1st  and  3d,  and  2#=40. 


( 

2.  Given   I  x  —  y       —  4      to  find  x,  y  and  z. 
I***      =  6  j 


Add  all  three,  and  3#=3  6         or         #=12. 

{    x—y-z=  6  I 
3.  Given   <  3y  —  x  —  z=12      to  find  a?,  y  and  z 


86  ELEMENTS  OF  ALGEBRA. 

Assume  x-\-y-\-z=s.     Add  this  equation  to  each  of  the  given 
equations,  and  we  then  have 

2#=  6+5,         (A) 

4</=12+5,         (B) 

8z=24+5.         (C) 
Multiply  (.#)  by  4,  and  (B)  by  2,  and  we  have 

8#=24+4s, 


82=24+  s. 

By  addition,        85=3  X  24-}- -7s.     Or  5=72. 
Put  this  value  of  s  in  equation  (.#)  and  we  shall  have 
22=6+72.      Or     #=3+36=39,   &c. 


4.  Given    #+£T/=100,    3/+|z=100,    z+4#=100  to  find 
ar,  t/,  and  z. 

Put   «=100.  Ans.     #=64,   y=72,  and  z=84. 

r  =  10 


5    Given   • 


u-\-v-\-y+z=l2 


to  find  the  value  of  each. 


v+#+t/+2'=14 
Here  are  Jive  letters  and  five  equations.     Each  letter  has  the 
same  coefficient,  one  understood.     Each  equation  has  4  letters, 
z  is  wanting  in  the  1st  equation,  y  in  the  2d,  <fec. 
Now  assume  w+v+#+?/+;z=5. 
Then         5— z=10         (£\ 


s  —  #=12 
s  —  v=13 
s—  w=14 

Add,  and        5s—  -5=60         Or  5=15. 
Put  this  value  of  s  in  equation  (.#),  and  z=5,  &c. 

6.  Given  #+y=a,   #+z=&,  y+z=c. 
Add  the  1st  and  2d,  and  from  the  sum  subtract  the  3d. 
Jlns.  #=£ 


EQUATIONS 


8? 


Given 


8.  Given 


O.  Given 


1O.  Given  - 


2a—ti+yf*  I 

3y=u-}-x-\-z  I  to  find  the  value  of  w,  x,  y, 
4z=u+x+y  [      and  z. 
u=x— 14      J 

j.  w=26,   #=40,  y=30,  z=24. 

2  f  #=24. 
7                         rfns.  J  y=60. 

3  z=120- 


f  ar=ilT«- 

•   \   2/  =  T5T« 
I  ^-T7i« 

x=20 


x-\-a= 


y  —  2z=40 
4y  —  x-\-3z=35 
t=l3 


—  u—  49 


(Art.  54.)  Problems  producing  simple  equations  involving  two 
or  more  unknown  quantities. 

1.  Find  three  numbers  such,  that  the  product  of  the  1st  and 
2d,  shall  be  600 ;  the  product  of  the  1st  and  3d,  300  ;  and   the 
product  of  the  2d  and  3d,  200. 

rfns.  The  numbers  are,  30,  20,  and  10. 

2.  Find  three  numbers,  such  that  the^rstf  with  5  the  sum  of 
the  second  and  third  shall  be  120  ;  the  second  with  |  the  differ- 
ence of  the  third  and  first  shall  be  70  ;  and  the  sum  of  the  three 
numbers  shall  be  190.  Ans.  50  ;  65  ;  75. 

3.  A  certain  sum  of  money  was  to  be  divided  among  three 
persons,  .#,  B,  and  C,  so  that  »#'s  share  exceeded  j  of  the  shares 
of  B  and  C  by  $120  ;  also  the  share  of  B  exceeded  f  of  the 
shares  of  A  and  C  by  $120;  and  the  share  of  C,  likewise,  ex- 
ceeded f  of  the  shares  of  *#  and  B  by  $120.     What  was  each 
person's  share? 

tins.  JT*  share,  $600 ;  £'s,  480  ;  and  C"s  360. 


88  ELEMENTS  OF  ALGEBRA. 

4.  Jl  and  B,  working  at  a  job,  can  earn  $40  in  6  days  ;  A  and 
C  together  can  earn  $54  in  9  days;  and  B  and  C$80  in  15 
days.  What  can  each  person  alone  earn  in  a  day? 

Let  A  earn  x,  B  earn  y,  and  C  earn  z  dollars  per  day,  then, 
By  the  question,  6x-J-  6y=40 
9.2=54 


Dividing  the  equations  by  the  coefficients  of  the  unknown 
quantities,  we  have,  ,      _AO 

- 


See  Problem  6.     (Art.  53.) 

A  man  has  4  sons.  The  sum  of  the  ages  of  the  first,  second 
and  third  is  18  years  ;  the  sum  of  the  ages  of  the  first,  second 
and  fourth  is  16  years  ;  the  sum  of  the  ages  of  the  first,  third 
and  fourth  is  14  years  ;  the  sum  of  the  ages  of  the  second,  third 
and  fourth  is  12  years.  What  are  their  ages?  See  Problem  5. 
(Art.  53.)  rfns.  Their  ages  are,  8,  6,  4,  2. 

5.  .#,  B  and  C  sat  down  to  play,  each  one  with  a  certain  num- 
ber of  shillings  ;  Jl  loses  to  B  and  C  as  many  shillings  as  each 
of  them  has.     Next  B  loses  to  Jl  and  C  as  many  as  each  of 
them  now  has.     Lastly,  C  loses  to  Jl  and  B  as  many  as  each  of 
them  now  has.     After  all,  each  one  of  them  has  16  shillings. 
How  much  did  each  one  gain  or  lose  ? 

Let  x=  the  number  of  shillings  Jl  had  at  first 
y—  B's  shillings,  and 
z—  C's  shillings. 

Then,  by  resolving  the  problem,  we  shall  find  #=26,  y=14 
and  z=S.  Therefore,  Jl  lost  10  shillings,  B  gained  2,  and  (78. 

N.  B.  When  the  equations  are  found,  divide  the  1st  by  4,  the 
2d  by  2,  and  then  compare  them  with  Ex.  3.  (Art.  53.) 

6.  A  gentleman  left  a  sum  of  money  to  be  divided  among  four 
servants,  so  that  the  share  of  the  first  was  £  the  sum  of  the 
shares  of  the  other  three  ;  the  share  of  the  second,  |  of  the  sum 
of  the  other  three  ;  and  the  share  of  the  third,  k  the  sum  of  the 


EQUATIONS.  89 

other  three ;  and  it  was  found  that  the  share  of  the  last  was  14 
dollars  less  than  that  of  the  first.  What  was  the  amount  of 
money  divided,  and  the  shares  of  each  respectively  1 

Ans.  The  sum  was  $120 ;  the  shares  40,  30,  24  and  26. 

Observe  Prob.  7.  (Art.  53,)  in  connection  with  this  problem. 

7.  A  jockey  has  two  horses,  and  two  saddles  which  are  worth 

15  and  10  dollars,  respectively.     Now  if  the  better  saddle  be  put 
on  the  better  horse,  the  value  of  the  better  horse  and  saddle 
would  be  worth  £  of  the  other  horse  and  saddle.     But  if  the 
better  saddle  be  put  on  the  poorer  horse,  and  the  poorer  saddle 
on  the  better  horse,  the  value  of  the  better  horse  and  saddle  is 
worth  once  and  f^  the  value  of  the  other.     Required  the  worth 
of  each  horse  ?  Ans.  65  and  50  dollars. 

8.  A  merchant  finds  that  if  he  mixes  sherry  and  brandy  in 
quantities  which  are  in  proportion  of  2  to  1,  he  can  sell  the  mix- 
ture at  78  shillings  per  dozen;  but  if  the  proportion  be  7  to  2  he 
can  sell  it  at  79  shillings  per  dozen.     Required  the  price  per 
dozen  of  the  sherry  and  of  the  brandy  ? 

Ans.  Sherry,  81s.     Brandy,  725. 

In  the  solution  of  this  question,  put  a=78.     Then  a-f-l=79. 

9.  Two  persons,  A  and  J9,  can  perform  a  piece  of  work  in 

16  days.     They  work  together  for  four  days,  when   A  being 
called  off,  B  is  left  to  finish  it,  which  he  does  in  36  days.     In 
what  time  would  each  do  it  separately  7 

Ans.  A  in  24  days ;  B  in  48  days. 

1O  What  fraction  is  that,  whose  numerator  being  doubled, 
and  denominator  increased  by  7,  the  value  becomes  f ;  but  the 
denominator  being  doubled,  and  the  numerator  increased  by  2, 
the  value  becomes  |?  Ans.  £. 

11.  Two  men  wishing  to  purchase  a  house  together,  valued 
at  240  (a)  dollars ;  says  A  to  J3,  if  you  will  lend  me  §  of  your 
money  I  can  purchase  the  house  alone ;  but  says  B  to  A,  if  you 
lend  me  J  of  yours,  I  can  purchase  the  house.  How  much 
money  had  each  of  them  ?  Ans.  A  had  $160.  B  $120. 

8 


90  ELEMENTS  OF  ALGEBRA. 

12.  It  is  required  to  divide  the  number  24  into  two  such  parts, 
that  the  quotient  of  the  greater  part  divided  by  the  less,  may  be 
to  the  quotient  of  the  less  part  divided  by  the  greater,  as  4  to  1 

rfns.  16  and  8. 

13.  A  certain  company  at  a  tavern,  when  they  came  to  settle 
their  reckoning,  found  that  had  there  been  4  more  in  company, 
they  might  have  paid  a  shilling  a-piece  less  than  they  did;  but 
that  if  there  had  been  3  fewer  in  company,  they  must  -have  paid 
a  shilling  a-piece  more  than  they  did.     What  then  was  the  num- 
ber of  persons  in  company,  and  what  did  each  pay  ? 

rfns.  24  persons,  each  paid  7s. 

14.  There  is  a  certain  number  consisting  of  two  places,  a  unit 
and  a  ten,  which  is  four  times  the  sum  of  its  digits,  and  if  27  be 
added  to  it,  the  digits  will  be  inverted.     What  is  the  number? 

Ans.  36. 

NOTE.  Undoubtedly  the  reader  has  learned  in  arithmetic  that 
numerals  have  a  specific  and  a  local  value,  and  every  remove 
from  the  unit  multiplies  by  10.  Hence,  if  x  represents  a  digit 
in  the  place  of  tens,  and  y  in  the  place  of  units,  the  number  must 
be  expressed  by  lQx-\-y.  A  number  consisting  of  three  places, 
with  x,  y  and  z  to  represent  the  digits,  must  be  expresset^  by 


15.  A  number  is  expressed  by  three  figures  ;  the  sum  of  these 
figures  is  1  1  ;  the  figure  in  the  place  of  units  is  double  that  in 
the  place  of  hundreds,  and  when  297  is  added  to  this  number, 
the  sum  obtained  is  expressed  by  the  figures  of  this  number  re- 
versed. What  is  the  number  ?  Jlns.  326. 

10.  Divide  the  number  90  into  three  parts,  so  that  twice 
the  first  part  increased  by  40,  three  times  the  second  part  in- 
creased by  20,  and  four  times  the  third  part  increased  by  10,  may 
be  all  equal  to  one  another. 

.  *ftns.  First  part  35,  second  30,  and  third  25. 

17.  A  person  who  possessed  $100,000  («,)  placed  the  greater 
part  of  it  out  at  5  per  cent,  interest,  and  the  other  part  at  4  per 


EQUATIONS.  91 

cent.     The  interest  which  lie  received  for  the  whole  amounted 
to  4640  (b~)  dollars.     Required  the  two  parts. 

Am.  64,000  and  36,000  dollars. 

General  Ansiver.  (1006 — 4«)  for  the  greater  part,  and  (5«— 
1006)  for  the  less. 

18.  A  person  put  out  a  certain  sum  of  money  at  interest  at  a 
certain  rate.     Another  person  put  out  $10,000  more,  at  a  rate  1 
per  cent,  higher,  and  received  an  income  of  $800  more.    A  third 
person  put  out  $15,000  more  than  the  first,  at  a  rate  2  per  cent, 
higher,  and  received  an  income  greater  hy  $1,500.     Required 
the  several  sums,  and  their  respective  rates  of  interest. 

Jlns.  Rates  4,  5  and  6  per  cent.     Capitals  $30,000,  $40,000 
and  $45,000. 

19.  A  widow  possessed   13,000  dollars,  which  she   divided 
into  two  parts  and  placed  them  at  interest,  in  such  a  mannei,  that 
the  incomes  from  them  were  equal.     If  she  had  put  out  the  first 
portion  at  the  same  rate  as  the  second,  she  would  have  drawn 
for  this  part  360  dollars  interest,  and  if  she  had  placed  the  se- 
cond out  at  the  same  rate  as  the  first,  she  would  have  drawn  for  it 
490  dollars  interest.     What  were  the  two  rates  of  interest  ? 

Ans.  7  and  6  per  cent.* 

20.  There  are  three  persons,  .#,  B  and  C,  whose  ages  are  as 
follows :  if  J5's  age  be  subtracted  from  ./2's,  the  difference  will 
be  C"s  age ;  if  five  times  J5's  age  and  twice  C"s  age,  be  added 
together,  and  from  their  sum  JFs  age  be  subtracted,  the  remain- 
der will  be  147.     The  sum  of  all  their  ages  is  96.     What  are 
their  ages  ?  Jlns.  £'s  48,  J5's  33,  <7's  15. 

21.  Find  what  each  of  three  persons,  ./?,  B  and  (7,  is  worth, 
from  knowing,  1st,  that  what  Ji  is  worth  added  to  3  times  what 
B  and  C  are  worth, make  4700  dollars  ;  2d,  that  what  B  is  worth 
added  to  four  times  what  A  and  C  are  worth  make  5800  dollars; 
3d,  that  what  C  is  worth  added  to  five  times  what  Ji  and  B  are 
worth  make  6300  dollars. 

tins.  A  is  worth  500,  B  600,  C  800  dollars. 

tt*;e  brief  solution  to  these  two  problems,  18  and  19,  in  Key  to  Algebra. 


92 


ELEMENTS  OF  ALGEBRA. 


Put  5=  the  sum  that  v?,  B  and  C  are  worth,  to  make  an  aux- 
iliary equation. 

22.  Find  what  each  of  three  persons,  «/?,  B,  C,  is  worth, 
knowing,  1st,  that  what  Jl  is  worth  added  to  /  times  whauE?  and 
C  are  worth  is  equal  to  p;  2d,  that  what  B  is  worth  added  to  m 
times  what  Ji  and  C  are  worth  is  equal  to  q;  3d,  that  what  C  is 
worth  added  to  n  times  what  Ji  and  B  are  worth  is  equal  to  r. 

Let     #=r#'s  capital,   y=B<>81   and   2  =  C"s. 
Then      x-^ly-^-lz—p^    is  the  first  equation. 

Assume  x-{-  y+  *=*•  Multiply  this  equation  by  /  and 
subtract  the  former,  and  (/  —  \}x=ls  —  p. 


Is — p 


r>  •     -i  *•  ms — *7 

By  a  similar  operation,      y= J 

And, 


m — 1 

ns — r 


(A) 


By  addiuon, 


ns  —  r 


This  equation  may  take  the  following  form  : 

=(-1 


+-^+- 


Now  as  the  terms  in  parenthesis  are  fully  determined,  of 
known  value,  we  may  represent  the  first  by  a,  the  second  by  bt 
and  this  last  form  becomes 

s=«5 — b 

By  transposition,  &c.      (a — l)s=6 

b 


Therefore 


*= 


a  —  1 


This  known  value  of  *  put  in  each  of  the  equations  marked 
(A),  and  the  values  of  x,  y  and  z  will  be  theoretically  deter- 
mined. 


EQUATIONS.  93 

23.  Three  brothers  made  a  purchase  of  $2000  (a;)  the  first 
wanted  in  addition  to  his  own  money  £  the  money  of  the  second, 
the  second  wanted  in  addition  to  his  own  3  of  the  money  of  the 
third,  and  the  third  required  in  addition  to  his  own  £  of  tho 
money  of  the  first.          How  much  money  had  each  ? 

Jlns.  1st  had,  $1280;  2d,  $1440;  and  the  3d,  $1680. 
Gen.  Jlns.  1st  had  if  a;  2d,  if  a;  and  the  3d  \  \a. 
See  Prob.  6.  (Art.  53.) 

24.  Some  hours  after  a  courier  had  been  sent  from  A  to  B, 
which  are  147  miles  distant,  a  second  was  sent,  who  wished  to 
overtake  him  just  as  he  entered  B,  and  to  accomplish  this  he 
must  perform  the  journey  in  28  hours  less  time  than  the  first  did 
Now  the  time  that  the  first  travels  17  miles  added  to  the  time  the 
second  travels  56  miles  is  13f  hours.     How  many  miles  does 
each  go  per  hour  ?  Jlns.  1st  3,  the  2d,  7  miles  per  hour. 

25.  There  are  two  numbers,  such  that  £  the  greater  added  to 
I  the  lesser,  is  13 ;  and  if  £  the  lesser  is  taken  from  5  the  greater, 
the  remainder  is  nothing.     Required  the  numbers. 

rfns.  18  and  12. 

26.  Find  three  numbers  of  such  magnitude,  that  the  1st  with 
the  5  sum  of  the  other  two,  the  second  with  £  of  the  other  two, 
and  the  third  with  *  of  the  other  two,  may  be  the  same,  and 
amount  to  51  in  each  case.  J%ns.  15,  33,  and  39. 

27.  Jl  said  to  B  and  Ct  "  Give  me,  each  of  you,  4  of  your 
sheep,  and  I  shall  have  4  more  than  you  will  have  left."     B  said 
to  A  and  C,  "  If  each  of  you  will  give  me  4  of  your  sheep,  I 
shall  have  twice  as  many  as  you  will  have  left."     C  then  said 
to  A  and  B,  "  Each  of  you  give  me  4  of  your  sheep,  and  I  shall 
have  three  times  as  many  as  you  will  have  left."     How  many 
had  each  ?  rfns.  Ji  6,  B  8,  and  C  10. 

28.  What  fraction  is  that,  to  the  numerator  of  wnich  if  1  be 
added,  the  fraction  will  be  j :  but  if  1  be  added  to  the  denomina- 
tor, the  fraction  will  be  ?  1  dns.  T4T. 

29.  What  fraction  is  that,  to  the  numerator  of  which  if  2  be 


U4  ELEMENTS  OF  ALGEBRA. 

added,  the  fraction  will  be  f  ;  but  if  2  be  added  to  the  denomina- 
tor, the  fraction  will  be  }  ?  Ans.  f . 

SO.  \Vhat  fraction  is  that  whose  numerator  being  doubled,  and 
its  denominator  increased  by  7,  the  value  becomes  f ;  but  the  do- 
nominator  being  doubled,  and  the  numerator  increased  by  2,  the 
value  becomes  |  ?  Ans.  J. 

31.  If  A  give  B  $5  of  his  money,  B  will  have  twice  as  much 
money  as  A  has  left;  and  if  B  give  A  $5,  A  will  have  thrice  as 
much  as  B  has  left.     How  much  had  each  ? 

Ans.  .tf  $13,  and  .#$11. 

32.  A  corn  factor  mixes  wheat  flour,  which  cost  him  10  shil- 
lings per  bushel,  with  barley  flour,  which  cost  4  shillings  per 
bushel,  in  such  proportion  as  to  gain  43f  per  cent,  by  selling  the 
mixture  at  11  shillings  per  bushel.     Required  the  proportion. 

Am.  The  proportion  is  14  bushels  of  wheat  flour  to  9  of 
barley. 

33.  There  is  a  number  consisting  of  two  digits,  which  num- 
ber divided  by  5  gives  a  certain  quotient  and  a  remainder  of  one, 
and  the  same  number  divided  by  8  gives  another  quotient  and  a 
remainder  of  one..     Now  the  quotient  obtained  by  dividing  by  5 
is  double  of  the  value  of  the  digit  in  the  ten's  place,  and  the  quo- 
tient obtained  by  dividing  by  8  is  equal  to  5  times  the  unit  digit. 
What  is  the  number  ?  Ans.  41. 

Interpretation  of  negative  values  resulting  from  the  solution 

of  equations. 

(Art.  55.)  The  resolution  of  proper  equations  drawn  from 
problems  not  only  reveal, the  numeral  result,  but  improper  enun- 
ciation by  the  change  of  signs.  Or  the  signs  being  true  algebraic 
language,  they  will  point  out  errors  in  relation  to  terms  in  com- 
mon language,  as  the  following  examples  will  illustrate : 

1.  The  sum  of  two  numbers  is  120,  and  their  difference  is 
100  ;  what  are  the  numbers  ? 

Let  x  be  the  greater  and  y  the  less.     Then 
a?+y=120         (1) 
a^=160         (2) 
The  solution  gives  #=140,  and  ?/— — 20. 


EQUATIONS.  95 

Here  it  appears  that  one  of  the  numbers  is  greater  than  the 
sum  given  in  the  enunciation,  yet  the  sum  of  x  and  y,  in  the  al- 
gebraic sense,  is  120. 

There  is  no  such  abstract  number  as  — 20,  and  when  minus 
appears  it  is  only  relative  or  opposite  in  direction  or  condition 
to  plus,  and  the  problem  is  susceptible  of  interpretation  in  an  al- 
gebraic sense,  but  not  in  a  definite  arithmetical  sense. 

Indeed  we  might  have  determined  this  at  once  by  a  considera- 
tion of  the  problem,  for  the  difference  of  the  two  numbers  is 
given,  greater  than  their  sum.  But  we  can  form  a  problem,  an 
algebraic  (not  an  abstract)  problem  that  will  exactly  correspond 
with  these  conditions,  thus : 

The  joint  property  of  two  men  amounts  to  120  dollars,  and 
one  of  them  is  worth  160  dollars  more  than  the  other.  What 
amount  of  property  does  each  possess  ? 

The  answer  must  be  -f-140  and  — 20  dollars  ;  but  there  is  no 
such  thing  as  minus  $20  in  the  abstract ;  it  must  be  interpreted 
debt,  an  opposite  term  to  positive  money  in  hand. 

2.  Two  men,  A  and  B,  commenced  trade  at  the  same  time ; 
Ji  had  3  times  as  much  money  as  B,  and  continuing  in  trade,  Jl 
gains  400  dollars,  and  J9  150  dollars;  now  A  has  twice  as  much 
money  as  B.  How  much  did  each  have  at  first? 

Without  any  special  consideration  of  the  question,  it  implies 
that  both  had  money,  and  asks  how  much.  But  on  resolving  the 
question  with  x  to  represent  ^?'s  money,  and  y  B's,  we  find 

#=—300 
And          y= — 100  dollars. 

That  is,  they  had  no  money,  and  the  minus  sign  in  this  case 
indicates  debt;  and  the  solution  not  only  reveals  the  numerical 
values,  but  the  true  conditions  of  the  problem,  and  points  out  the 
necessary  corrections  of  language  to  correspond  to  an  arithmeti- 
cal sense,  thus : 

A  is  three  times  as  much  in  debt  as  B ;  but  A  gains  400  dol- 
lars, and  B  150 ;  now  A  has  twice  as  much  money  as  B.  How 
much  were  each  in  debt  ? 

As  the  enunciation  of  this  problem  corresponds  with  the  real 


96  ELEMENTS  OF  ALGEBRA. 

circumstance  of  the  case,  we  can  resolve  the  problem  without  a 
minus  sign  in  the  result.     Thus : 

Let         z=  B'B  debt,     then       3x=  J?s  debt 

150— x—  J?'s  money,   400 — 3x=  .#'s  money 
Per  question,     400 — 3#-=300 — 2x.         Or       a?=100. 

3.  What  number  is  that  whose  fourth  part  exceeds  its  third 
part  by  12  ?  J2ns.  — 144. 

But  there  is  no  such  abstract  number  as  — 144,  and  we  cannot 
interpret  this  as  debt.  It  points  out  error  or  impossibility,  and 
by  returning  to  the  question  we  perceive  that  a  fourth  part  of  any 
number  whatever  cannot  exceed  its  third  part;  it  must  be,  its  third 
part  exceeds  its  fourth  part  by  12,  and  this  enunciation  gives  the 
positive  number,  144.  Thus  do  equations  rectify  subordinate 
errors,  and  point  out  special  conditions. 

4L  A  man  when  he  was  married  was  30  years  old,  and  his 
wife  15.  How  mapy  years  must  elapse  before  his  age  will  be 
three  times  the  age  of  his  wife? 

Jlns.  The  question  is  incorrectly  enunciated  ;  7£  years  before 
the  marriage,  not  after,  their  ages  bore  the  specified  relation. 

5.  A  man  worked  7  days,  and  had  his  son  with  him  3  days  , 
and  received  for  wages  22  shillings.     He  afterwards  worked  5 
days,  and  had  his  son  with  him  one  day,  and  received  for  wages 
18  shillings.     What  were  his  daily  wages,  and  the  daily  wages- 
of  his  son  ? 

Jfns.  The  father  received  4  shillings  per  day,  and  paid  2  shil- 
lings for  his  son's  board. 

6.  A  man  worked  for  a  person  ten  days,  having  his  wife  with 
him  8  days,  and  his  son  6  days,  and  he  received  $10.30  as  com- 
pensation for  all  three;  at  another  time  he  wrought  12  days,  his 
wife  10  days,  and  son  4  days,  and  he  received  $-13.20;  at  an- 
other time  he  wrought  15  days,  his  wife  10  days,  and  his  son  12 
days,  at  the  same  rates  as  before,  and  he  received  $13.85.    What 
were  the  daily  wages  of  each  ? 

rfns.  The  husband  75  cts.,  wife  50  cts.  The  son  20  cts.  ex- 
pense per  day. 


EQUATIONS.  97 

7.  A  man  wrought  10  days  for  his  neighbor,  his  wife  4  days, 
and  son  3  days,  and  received  $11.50 ;  at  another  time  he  served 
9  days,  his  wife  8  days,  and  his  son  6  days,  at  the  same  rates  as 
before,  and  received  $12.00  ;  a  third  time  he  served  7  days,  his 
wife  6  days,  and  his  son  4  days,  at  the  same  rates  as  before,  and 
he  received  $9.00.     What  were  the  daily  wages  of  each  ? 

*fins.  Husband's  wages,$  1.00;  wife  0 ;  son  50  cts. 

8.  What  fraction  is  that  which  becomes  f  when  one  is  added 
to  its  numerator,  and  becomes  |  when  1  is  added  to  its  denomi- 
nator ? 

Jlns.  In  an  arithmetical  sense,  there  is  no  such  fraction.  The 
algebraic  expression,  ~y|,  will  give  the  required  results. 

(Art.  58.)  By  the  aid  of  algebraical  equations,  we  are  enabled 
not  only  to  resolve  problems  and  point  out  defects  or  errors  in 
their  enunciation,  as  in  the  last  article,  but  we  are  also  enabled 
to  demonstrate  theorems,  and  elucidate  many  philosophical  truths. 
The  following  are  examples  : 

Theorem  1.  It  is  required  to  demonstrate,  that  the  half  sum 
plus  half  the  difference  of  two  quantities  give  the  greater  of  the 
two,  and  the  half  sum  minus  the  half  difference  give  the  less. 

Let  x=  the  greater  number,         y=  the  less, 

s=  their  sum,  d—  their  difference. 

Then  x+y=  s  (A] 

And    x—y=d  (B) 

By  addition,  2a?=  s-f-  d 

Or  x=*s-}-zd 

Subtract  (S)  from  (.#)  and  divide  by  2,  and  we  have 
y=%s — |c? 

These  last  two  equations,  which  are  manifestly  true,  demon- 
strate the  theorem. 

Theorem  2.  Four  times  the  product  of  any  two  numbers,  is 
equal  to  the  square  of  their  sum,  diminished  by  the  square  of 
their  difference. 

9 


98  ELEMENTS  OF  ALGEBRA 

Let  x=  the  greater  number,  and  y=  the  less,  as  in  the  lasi 
theorem.  2x=s  -\-d 

And         2=s  —  d 


By  multiplication  4xy=s2  —  d2  a  demonstration  of  the 
theorem. 

Many  other  theorems  are  demonstrable  by  algebra,  but  we  de- 
fer them  for  the  present,  as  some  of  them  involve  quadratic  equa- 
tions, which  have  not  yet  been  investigated  ;  and  we  close  the 
subject  of  simple  equations  by  the  following  quite  general  prob- 
lem in  relation  to  space,  time  and  motion. 

To  present  it  at  first,  in  the  most  simple  and  practical  manner, 
let  us  suppose 

Two  couriers,  A  and  B,  100  miles  asunder  on  the  same  road 
set  out  to  meet  each  other,  A  going  6  miles  per  hour  and  B  4. 
How  many  hours  must  elapse  before  they  meet,  and  how  far 
will  each  travel? 

Let  x=  Jl's  distance,  y  —  j#'s,  and  t—  the  time. 
Then       ^-}-?/=100    '     (1) 

As  the  miles  per  hour  multiplied  by  the  hours  must  give  the 
distance  each  traveled,  therefore, 

x=6t     and     y=U         (2) 

Substitute  these  values  in  equation  (1)  and 
(6+4)f=  100 

1  rjn 

Therefore,          *=r  (3) 


100X6  100X4 

And        *=8l=-  2/=4/= 


From  equation  (3,)  we  learn  that  the  time  elapsed  before  the 
couriers  met  was  the  whole  distance  divided  by  their  joint  mo- 
tion per  hour,  a  result  in  perfect  accordance  with  reason.  From 
equations  (4,)  we  perceive  that  the  distance  each  must  travel  is 
.he  whole  distance  asunder  multiplied  by  their  respective  mo- 
tions and  divided  by  the  sum  of  their  hourly  motions. 

Now  let  us  suppose  the  couriers  start  as  before,  but  travel  in 
the  same  direction,  the,  one  in  pursuit  of  the  other.  B  having 


EQUATIONS.  99 

IQO  miles  the  start,  traveling  four  miles  per  hour,  pursued  by 

A,  traveling  6  miles  per  hour.     How  many  hours  must  elapse 
before  they  come  together,  and  what  distance  must  each  travel? 

Take  the  same  notation  as  before. 

Then  x  —  y=100  (1.)     As  A  must  travel  100  miles  more  than 

B.  But  equations  (2,)  that  is,  x=Qt  and  y=±t,  are  true  under 
all  circumstances. 

Then         (6—4)*=  100 

And  i 


The  result  in  this  case  is  as  obvious  as  an  axiom.  A  has  100 
miles  to  gain,  and  he  gains  2  miles  per  hour,  it  will  therefore  re- 
quire 50  hours. 

But  it  is  the  precise  form  that  we  wish  to  observe.  It  is  the 
fact  that  the  given  distance  divided  by  the  difference  of  their  ma 
tions  gives  the  time,  and  their  respective  distances  must  be  this 
time  multiplied  by  their  respective  rates  of  motion. 

Now  the  smaller  the  difference  between  their  motions,  the 
longer  the  time  before  one  overtakes  the  other  ;  when  the  differ 
ence  is  very  small,  the  time  will  be  very  great  ;  when  the  differ 
ence  is  nothing,  the  time  will  be  infinitely  great  ;  and  this  is  in 
perfect  accordance  with  reason  ;  for  when  they  travel  equally 
fast  one  cannot  gain  on  the  other,  and  they  can  never  come  to- 
gether. 

If  the  foremost  courier  travels  faster  than  the  other,  they  must 
all  the  while  become  more  and  more  asunder  ;  and  if  they  have 
ever  been  together  it  was  preceding  their  departure  from  the 
points  designated,  and  in  an  opposite  direction  from  the  one  they 
are  traveling,  and  would  be  pointed  out  by  a  negative  result. 

(Art.  59.)  Let  us  now  make  the  problem  general. 

Two  couriers,  A  and  B,  d  miles  asunder  on  the  same  road, 
set  out  to  meet  each  other;  A  going  a  miles  per  hour,  B  going 
b  miles  per  hour.  How  many  hours  must  elapse  before  they 
meet,  and  how  far  will  each  travel? 


100  ELEMENTS  OF  ALGEBRA. 

Taking  the  same  notation  as  in  the  particular  case, 
Let  x—  *#'s  distance,   y=  .5's,    and   /=  the  time. 

Then          x-\-y=d     (I)          x=at          y—bt  (2) 

Therefore         (a-\-b)t=d  Or  t==i~TJ)        (3) 

ad  bd 

If  a=b,  then  x=kd  and  y=%.d.  A  result  perfectly  obvious, 
the  rates  being  equal.  Each  courier  must  pass  over  one  half 
the  distance  before  meeting. 

If  «=0         #===     ,  .  =0     and     y=-—=d.      That    is,   one 
0+6  0 

will  be  at  rest,  and  the  other  will  pass  over  the  whole  distance. 

(Art.  60.)  Now  let  us  consider  the  other  case,  in  which  one 
courier  pursues  the  other,  starting  at  the  same  time  from  dif- 
ferent points. 

Let  the  line  CD  represent  the  space  the  couriers  are  asunder 
when  the  pursuit  commences,  and  the  point  E  where  they  come 
together.  C  D  E 

j  j  j 

The  direction  from  C  towards  D  we  call  plus,  the  other  direc- 
lion  will  therefore  be  minus. 

Now  as  in  the  2d  example,  (Art.  58.) 

Put        x  =  CE  =^'s  distance 
y=DE=JS1s  distance 
Then        x—y=CD=d  (1) 

As  before,  let        t=    the  time.     Then 

Therefore     at — bt=d 

And  t=^  (3) 

For  the  distances  we  have 

ad       ,  .  t/tt  .  N 

x= (4)       and      y= ,  (o) 

0—6     v  '  9    "—h 


EQUATIONS.  101 

By  an  examination  of  these  equations,  it  will  be  perceived  fazA 
x  and  y  will  be  equal  when  a  is  equal  to  b,  yet  d  still  exists  as 
a  difference  between  them.  This  is  in  consequence  of  x  and  y 
in  that  case  being  so  very  great  that  d  is  lost  in  comparison.  So 
all  values  are  great  or  small  only  in  comparison  with  others  or 
with  our  scale  of  measure. 

To  make  this  clear,  let  us  suppose  two  numbers  differ  by  one 
and  if  the  numbers  are  small,  the  difference  may  be  regarded  as 
considerable;  if  large,  more  inconsiderable;  if  still  larger,  still 
more  inconsiderable,  &c.  If  the  numbers  or  quantities  be  infin- 
itely great,  the  comparative  small  quantity  may  be  rejected. 
Thus : 

5  and  6  differ  by  1,  and  their  relation  is  as  1  to  1.2. 

Also,  50  and  51  differ  by  1,  and  their  relation  is  as  1  to  1.02. 
500  to  501  are  as  1  to  1.002,  <fcc.  The  relation  becoming  nearer 
and  nearer  equality  as  the  numbers  become  larger,  and  when  the 
numbers  become  infinitely  great  the  difference  is  comparatively 
nothing. 

When  a=b     a — ft— 0     and    x~~ft      a  symbol  of  infinity. 

If  we  suppose  b  greater  than  a,  a — b  will  become  negative,  and 
as  x  and  y  refer  to  the  same  point,  that  point  must  then  be  in  the 
backward  direction  from  that  we  suppose  the  couriers  are  mo- 
ving, and  will  show  how  far  they  have  traveled  since  that  event. 

If  in  the  equations  (3),  (4)  and  (5),  c?=0,  and  at  the  same 
time  a=bj  then  we  shall  have  /=-,  #=-  and  2/=~  '•  which 

shows  that  -  is  a  symbol  of  indetermination,  it  being  equal  to 

several  quantities  at  the  same  time.  If  e?=0  the  two  couriers 
were  together  at  commencement ;  and  if  they  travel  in  the  same 
direction,  and  equally  fast,  they  will  be  together  all  the  while, 
and  the  distances  represented  by  x  and  y  will  be  equal,  and  of 

all  possible  values.     Hence  -  may  be  taken  of  any  value  what- 


102  CLEMENTS  OF  ALGEBRA 

evsr^  amV-may  Ve'-vntidMo  take  a  particular  value,  to  correspond 
to  any  other  circumstance  or  condition.* 

APPLICATION. 

(Art.  61.)  We  have  hitherto  considered  CD  a  right  line;  but 
the  equations  would  be  equally  true,  if  we  consider  CD  to  be 
curved,  and  indeed  we  can  conceive  the  line  CD  to  wind  about 
a  perfect  circle  just  forming  its  circumference,  and  the  point  E 
upon  the  circle,  CE  being  a  little  more  than  one  circumference. 

This  being  understood,  Equation  (3,)  (Art.  60,)  gives  us  a  so- 
lution to  the  following  problems. 

1.  The  hour  and  minute  hands  of  a  clock  are  together  at  12 
o'clock.  When  are  they  next  together  ? 

*  The  26th  equation  (Art.  40),  if  resolved  in  the  briefest  manner,  will  show 

0 
the  influence  of  the  factor   rr.     In  the  equation  referred  to,  add  30  to  both 

members  and  divide  the  numerator  of  the  second  member  by  its  denominator, 

5^4-5 
and  we  have   —  -  --  [-1=6.     Drop  1,  and  divide  both  members  by  5,  we  then 


x4-\ 
have    —  —  =1,       or      x-\-l=x-}-2.     Hence    1=2,   a  manifest  absurdity 


But  all  our  operations,  yea,  and  all  our  reasonings  have  been  correct,  but 
we  did  not  pay  sufficient  attention  to  dividing  the  numerator  by  the  denomi- 

nator, which  was   -7  -  —  .     Taking  6  for  the  quotient,  which  it  would  be  in 

(x  —  t) 

every  case  except  when  x  —  2=0,  leads  to  the  absurdity  ;  which  absurdity, 
in  turn,  shows  that  x  —  2=0,  or  #=2. 

As  another  illustration  of  the  influence  of  this  symbol,  take  the  identical 
equation    100=100,   or  any  other  similar  one. 
This  is  the  same  as  96-f-4=96-f-4 

Transposing,  .     4  —  4=96  —  96 

Resolving  into  factors,  1(4—  4)  =24  (4—  4) 

Dividing  by  the  common  factor,  and     1=24  ; 

0  1 

But,  to  restore  equality,   -    in  this  case  must  equal    —  ,  or  24. 

0 
Hence  we  perceive  that    -  is  indeterminate,  in  the  abstract,  but  may  be  ren- 

dered definite  in  particular  cases. 


EQUATIONS.  103 

By  the  equation,   t— j-  this  problem  and  all  others  like  it 

are  already  resolved.     All  we  have  to  do  is  to  determine   the 
values  of  </,  «,  and  b. 

There  are  12  spaces  (hours)  round  the  dial  plate  of  a  clock ; 
hence  d  may  represent  12.  a  and  b  are  the  relative  motions  ol 
the  hands,  a  moves  12  spaces  or  entirely  round  the  dial  plate 
while  b  moves  one  space.  Hence  a=12,  6=1,  and  a — 6=11. 

Consequently,         t=\\—lh.  5m.  27T\s. 

Again.  We  may  demand  what  time  the  hour  and  minute 
hands  of  a  clock  are  together  between  3  and  4. 

From  12  o'clock  to  past  3  o'clock  there  are  3  revolutions  to 

q  v/  i  o 

pass  over  in  place  of  one,  and  the  solution  is  therefore  t— 

and  so  on  for  any  other  hour. 

2.  WJiat  time  between  two  and  three  o'clock  will  the  hour 
and  minute  hands  of  a  clock  make,  right  angles  with  each  other? 

Here  the  space  that  the  one  courier  must  gain  on  the  other  is 
two  revolutions  and  a  quarter,  or  21  d. 

Hence      t=\\Wk=\\ X  J=H=2h.  27T3Tm. 

3.  What  time  between  5  and  6  will  the  two  hands  of  a  clock 
make  a  right  line  ? 

Here  one  courier  must  gain  5 £  revolutions,  or  d  in  the  equa- 
tion must  be  multiplied  by  5£  =  y . 

Hence,         t=\\  X  V  =6h. 

That  is,  the  hands  make  a  right  line  at  6  o'clock,  a  result  man- 
ifestly tri'3. 

This  simple  equation  enables  us  to  determine  the  exact  time 
when  the  two  hands  of  a  clock  shall  be  in  any  given  position. 

We  may  apply  this  equation  to  a  large  circle,  as  well  as  to  a 
small  one;  it  may  apply  to  the  apparent  circular  course  of  the 
heavens,  as  well  as  to  a  dial  plate  of  a  clock ;  and  the  application 
is  equally  simple. 

The  circle  of  the  heavens,  like  all  other  circles,  is  divided  into 


104  ELEMENTS  OF  ALGEBRA. 

360  degrees  ;  and  the  sun  and  moon  apparently  follow  each  other 
like  two  couriers  round  the  circle. 

In  one  day  the  moon  moves  on  an  average  13°.  1764,  (divisions 
of  the  circle,)  and  the  sun  apparently  0°.  98565,  or  not  quite  one 
division  of  the  circle.  The  moon's  motion  being  most  rapid. 
corresponds  to  a  in  the  equation,  and  the  sun's  apparent  motion 
to  b.  Then  «—  &=13°.1764—  0.98565=12°.19075  ;  and  the 
time  required  for  one  courier  to  gain  on  the  other  the  required 

space,  in  this  case  a  revolution  of  360  degrees,  or        /—  —  .-= 

360  a~"b 

——  —  —  which  gives  29.5305887  days,  or  29  days,  12  hours, 
\Z.  19075 

44  minutes,  3  seconds,  which  is  the  mean  time  from  one  change 
of  the  moon  to  another,  called  a  synodic  revolution. 

These  relative  apparent  motions  of  the  sun  and  moon  round 
the  circular  arc  of  the  heavens,  are  very  frequently  compared  to 
the  motions  of  the  hour  and  minute  hands  of  a  clock  round  the 
dial  plate  ;  and  from  the  preceding  application  of  the  same  equa- 
tion we  see  how  truly. 

We  may  not  only  apply  this  equation  to  the  mean  motions  of 
the  sun  and  moon,  but  it  is  equally  applicable  to  the  mean  mo- 
tions of  any  two  planets  as  seen  from  the  sun.  To  appearance, 
the  two  planets  would  be  nothing  more  than  two  couriers  mo- 
ving in  a  circle,  the  one  in  pursuit  of  the  other,  and  the  time  be- 
tween two  intervals  of  coming  together,  (or  coming  in  conjunc- 
tion, as  it  is  commonly  expressed,)  will  be  invariably  represen- 
ted by  the  equation 


To  apply  this  to  the  motion  of  two  planets,  we  propose  the 
following  problem  : 

The  planet  Venus,  as  seen  from  the  sun,  describes  an  arc  of 
1°  36'/)er  day,  and  the  earth,  as  seen  from  the  same  point,  de- 
scribes an  arc  of  59'.  Jit  what  interval  of  time  will  these  two 
bodies  come  in  a  line  with  the  sun  on  the  same  side  ? 

Here  a=l°36'          6=59'          rf=360° 

Therefore,  a  —  6=37';         and  as  the  denominator  is 


EQUATIONS.  105 


minutes,  the  numerator  must  be  reduced  to  minutes  also  ;  hence 
the  equation  becomes 


360X60    , 

—  —  —  =583.8  days,  nearly. 


We  have  not  been  very  minute,  as  the  motions  of  the  planets 
are  not  perfectly  uniform,  and  the  actual  interval  between  succes 
sive  conjunctions  is  slightly  variable.  Hence  we  were  not  par- 
ticular  to  take  the  values  of  a  and  b  to  the  utmost  fraction.  A 
more  rigid  result  would  have  been  583.92  days.  Half  of  this 
time  is  the  interval  that  Venus  remains  a  morning  and  an  even- 
ing star. 

(Art.  63.)  This  equation,  as  simple  as  it  may  Ep'  er.r,  is  one 
practical  illustration  of  the  true  spirit  and  utility  oi  analysis  by 
algebra. 

The  principles  and  relations  of  time  and  motion  are  fixed  and 

d 
invariable,  and  the  equation  t=^—^    stands  conspicuously   as 

one  of  these  relations. 

If  t  can  be  determined  by  observation,  as  it  may  be  in  respect 
to  the  earth  and  the  superior  planets,  the  mean  daily  motions  of 
the  planets  can  be  determined  ;  as  f?=360°,  a=59'  08"  the  mean 
motion  of  the  earth,  and  suppose  b  the  motion  of  Mars,  for  ex- 
ample, to  be  unknown. 

When  unknown,  represent  it  by  x. 

Then       t=  -          or      at  —  tx—d. 
a  —  x 

at  —  d 

Therefore         x=  —  -.  —  . 
t 


106  ELEMENTS  OF  ALGEBRA. 

SECTION    III. 

INVOLUTION. 
CHAPTER  I. 

(Art.  64.)  Equations,  and  the  resolution  of  problems  producing 
equations,  do  not  always  result  in  the  first  powers  of  the  un- 
known terms,  but  different  powers  are  frequently  involved,  and 
therefore  it  is  necessary  to  investigate  methods  of  resolving  equa- 
tions containing  higher  powers  than  the  first ;  and  preparatory 
to  this  we  must  learn  involution  and  evolution  of  algebraic 
quantities. 

(Art.  65.)  Involution  is  the  method  of  raising  any  quantity  to 
a  given  p  v  =h.  Evolution  is  the  reverse  of  involution,  and  is 
the  method  of  Determining  what  quantity  raised  to  a  proposed 
power  will  produce  a  given  quantity. 

As  in  arithmetic,  involution  is  performed  by  multiplication,  and 
evolution  by  the  extraction  of  roots. 

The  first  power  is  the  root  or  quantity  itself; 

The  second  power,  commonly  called  the  square,  is  the  quan- 
tity multiplied  by  itself ; 

The  third  power  is  the  product  of  the  second  power  by  the 
quantity ; 

The  fourth  power  is  the  third  power  multiplied  into  the  quan- 
tity, &c. 

The  second  power  of  a  is     a  X#     or    az 

The  third  power  is  azXa     or     a3 

The  fourth  power  is  a3Xa      or     a4 

The  second  power  of  a1  is     a4X«4     or     a8 

The  third  power  of  a4  is        a8Xa4     or     a12 

The  nth  power  of  a4  has  the  exponent  4  repeated  n  times, 
or  a4".  Therefore,  to  raise  a  simple  literal  quantity  to  any 
power,  multiply  its  exponent  by  the  index  of  the  required 
power. 

Raise  x  to  the  5th  power.  The  exponent  is  1  understood, 
and  this  1  multiplied  by  5  gives  x5  for  the  5th  power. 


INVOLUTION.  107 

Raise  x3  to  the  4th  power  Am.     a?12. 

Raise  y1  to  the  third  power.  Ans.    y21. 

Raise  xn  to  the  6th  power.  Am.     x*n. 

Raise  xn  to  the  mth  power.  Am.     a?17*. 

Raise  ay?  to  the  3d  power.  A*is.     a*x*. 

Raise  ab*x4  to  the  2d  power.  .tfns.     «26V. 

R*.ise  c27/4  to  the  5th  power.  .tfns.     c10*/20. 

(Art.  66.)  By  the  definition  of  powers  the  second  power  is 
any  quantity  multiplied  by  itself;  hence  the  second  power  of 
ax  is  a2^2,  the  second  power  of  the  coefficient  a,  as  well  as  the 
other  quantity  x ;  but  a  may  be  a  numeral,  as  6#,  and  its  second 
power  is  36x*.  Hence,  to  raise  any  simple  quantity  to  any 
power,  raise  the  numeral  coefficient,  as  in  arithmetic,  to  the 
required  power,  and  annex  the  powers  of  the  given  literal 
quantities 

EXAMPLES. 

1.  Required  the  3d  power  of  3a;r2.  Am.     27«V. 

2.  Required  the  4th  power  of  |?/2.  Jins.     Ify8. 

3.  Required  the  3d  power  of  — 2x.  Ans.     — Sx3. 
4U  Required  the  4th  power  of  — 3x.  Jlns.     8 la;4. 

Observe,  that  by  the  rules  laid  down  for  multiplication,  the 
even  powers  of  minus  quantities  must  be  plus,  and  the  odd  powers 
minus. 


5.  Required  the  2d  power  of  — - — .  Ans.    - — ^-. 

6.  Required  the  6th  power  of  — — .  Ans. 


7.  Required  the  6th  power  of  -j— .  Ans      ^ — . 

^a?  x 

(Art.  67.)  The  powers  of  compound  quantities  are  raised  by 
the  mere  application  of  the  rule  for  compound  multiplication. 
(Art.  12.) 


108  ELEMENTS  OF  ALGEBRA. 

Let  a-\-b  be  raised  to  the  2d,  3d,  4th,  &c.  powers. 
a  +b 


tf+ab 
ab+b* 


2d  power  or  square,  az-}-2ab 
a+b 


3d  power  or  cube, 


The  4th  power,        a4+4a36-f  Ga*b2-{-4abs+b< 
a+b 


The  5th  power, 


&c. 


By  inspecting  the  result  of  each  product,  we  may  arrive  at 
general  principles,  according  to  which  any  power  of  a  binomial 
may  be  expressed,  without  the  labor  of  actual  multiplication. 
This  theorem  for  abbreviating  powers,  and  its  general  application 
to  both  powers  and  roots,  first  shown  by  Sir  Isaac  Newton,  has 
given  it  the  name  of  Newton's  binomial,  or  the  binomial  theorem. 

OBSERVATIONS. 

Observe  the  5th  power  :  a,  being  the  first,  is  called  the  leading 
term  ;  and  6,  the  second,  is  called  the  following  term.  The  sum 
of  the  exponents  of  the  two  letters  in  each  and  all  of  the  terms 
amount  to  the  index  of  the  power.  In  the  5th  power,  the  sura 


INVOLUTION.  109 

of  the  exponents  ol  a  and  b  is  5  ;  in  the  4th  power  it  is  4  ;  in 
the  10th  power  it  would  be  10,  &c.  In  the  2d  power  there  are 
three  terms;  in  the  3d  power  there  are  4  terms;  in  the  4th 
power  there  are  5  terms  ;  always  one  more  term  than  the  index 
of  the  power  denotes. 

The  2d  letter  does  not  appear  in  the  first  term  ;  the  1st  letter 
does  not  appear  in  the  last  term. 

The  highest  power  of  the  leading  term  is  the  index  of  the 
given  power,  and  the  powers  of  that  letter  decrease  by  one  from 
term  to  term.  The  second  letter  appears  in  the  2d  term,  and  its 
exponent  increases  by  one  from  term  to  term  as  the  exponent  of 
the  other  letter  decreases. 

The  8th  power  of  («+&)  is  indicated  thus:  (a-f-6)8.  When 
expanded,  its  literal  part,  (according  to  the  preceding  observa- 
tions) must  commence  with  a8,  and  the  sum  of  the  exponents  of 
every  term  amount  to  8,  and  they  will  stand  thus  :  a8,  a76,  «662 
«563,  a4b4,  «365,  a266,  ab\  b9. 

The  coefficients  are  not  so  obvious.  However,  we  observe 
that  the  coefficients  of  the  first  and  last  terms  must  be  unity. 
The  coefficients  of  the  terms  next  to  the  first  and  last  are  equal, 
and  the  same  as  the  index  of  the  power.  The  coefficients  in- 
crease to  the  middle  of  the  series,  and  then  decrease  in  the 
same  manner,  and  it  is  manifest  that  there  must  be  some  law  of 
connection  between  the  exponents  and  the  coefficients. 

By  inspecting  the  5th  power  of  a-f-#»  we  find  that  the  2d  co- 
efficient is  5,  and  the  3d  is  10. 


The  third  coefficient  is  the  2d,  multiplied  by  the  exponent  of 
ihe  leading  letter,  and  divided  by  the  exponent  of  the  second 
Letter  increased  by  unity. 

In  the  same  manner,  the  fourth  coefficient  is  the  third  multi- 
plied by  the  exponent  of  the  leading  letter,  and  divided  by  the 
exponent  of  the  second  letter  increased  by  unity,  and  so  on  from 
coefficient  to  coefficient. 


HO 

The  4th  coefficient  is 

The  5th  is 
The  last  is 

Now  let  us  expand 
For  the  1st  term  write 
For  the  2d  term  write 

For  the  3d, 


ELEMENTS  OF  ALGEURA. 

1  ft  S/  Q 


=iO 


=  1    understood. 


I 


a8 
Sa'b 

28«e&2 
56a563 


For  the  4th, 


For  the  5th, 


Now  as  the  exponents  of  a  and  6  are  equal,  we  have  arrived 
at  the  middle  of  the  power,  and  of  course  to  the  highest  coeffi- 
cient. The  coefficients  now  decrease  in  the  reverse  order  which 
they  increased. 

Hence  the  expanded  power  is 


Let  the  reader  observe,  that  the  exponent  of  6,  increased  by 
unity  is  always  equal  to  the  number  of  terms  from  the  beginning, 
or  from  the  left  of  the  power.  Thus,  bz  is  in  the  3d  term,  &c. 
Therefore  in  finding  the  coefficients  we  may  divide  by  the  num- 
ber of  terms  already  written,  in  place  of  the  exponents  of  the 
second  term  increased  by  unity. 

If  the  binomial  (a-\-b)  becomes  (a+1,)  that  is,  when  b  be- 
comes unity,  the  8th  power  becomes, 


Any  power  of  1  is  1,  and  1  as  a  factor  never  appears. 
If  a  becomes  1,  then  the  expanded  power  becomes, 


INVOLUTION.  Ill 

The  manner  of  arriving  at  these  results  is  to  represent  the  unit 
by  a  letter,  and  expand  the  simple  literal  terms,  and  afterwards 
substitute  their  values  in  the  result. 

(Art.  68.)  If  we  expand  (a  —  6)  in  place  of  (a-\-b^  the  expo- 
nents and  coefficients  will  be  precisely  the  same,  but  the  princi- 
ples of  multiplication  of  quantities  affected  by  different  signs 
will  give  the  minus  sign  to  the  second  and  to  every  alternate  term. 

Thus  the  6th  power  of  (a  —  b)  is 


(Art.  69.)  This  method  of  readily  expanding  the  powers  of  a 
binomial  quantity  is  one  application  of  the  "  binomial  theorem" 
and  it  was  thus  by  induction  and  by  observations  on  the  result 
of  particular  cases  that  the  theorem  was  established.  Its  rigid 
demonstration  is  somewhat  difficult,  but  its  application  is  simple 
and  useful. 

Its  most  general  form  may  arise  from  expanding  («-{-£>)". 

When  n=3,  we  can  readily  expand  it; 

When  »=4,  we  can  expand  it  ; 

When  n=  any  whole  positive  number,  we  can  expand  it. 

Now  let  us  operate  with  n  just  as  we  would  with  a  known 
number,  and  we  shall  have 

an-zb2,       &c. 


We  know  not  where  the  series  would  terminate  until  we 
know  the  value  of  n.  We  are  convinced  of  the  truth  of  the 
result  when  n  represents  any  positive  whole  number;  but  let  n 
be  negative  or  fractional,  and  we  are  not  so  sure  of  the  result. 
To  extend  it  to  such  cases  requires  deeper  investigation  and 
rigid  demonstration,  which  it  would  not  be  proper  to  go  into  at 
this  time.  We  shall,  therefore,  content  ourselves  with  some  of 
its  more  simple  applications. 

EXAMPLES. 

1.  Required  the  third  power  of  3o?+2y. 
We  cannot  well  expand  this  by  the  binomial  theorem,  because 
the  terms  are  not  simple  literal  quantities.     But  we  can  assume 
3a?=0     and     2y=b.     Then 
o-f-&     and 


112  ELEMENTS  OF  ALGEBRA. 

Now  to  return  to  the  values  of  a  and  &,  we  have, 


X  4i/2=36;r«/2. 


Hence     ( 

2.  Required  the  4th  power  of  2a2  —  3. 

Let   #=2a2    2/=3.     Then   expand    (a?  —  y}4,   and  return  tho 
values  of  x  and  y,  and  we  shall  find  the  result, 
16a8—  96a64-216a4—  216«2+81. 

3.  Required  the  cube  of  (a-{-b  -j-c+rf). 

As  we  can  operate  in  this  summary  manner  only  on  binomial 
quantities,  we  represent  a-\-b  by  x,  or  assume  x=a-\-b,  and 


Then 

Returning  the  values  of  x  and  y,  we  have 


Now  we  can  expand  by  the  binomial,  these  quantities  con 
tained  in  parenthesis. 

4.  Required  the  4th  power  of  2a-{-3x. 

Ans. 

5.  Expand  (xz+3y*)s. 


6.  Expand  (2az-\-ax)* 

7.  Expand   (#—  I)6. 


8.  Expand  (3*—  -5)3  rfns.     27^—135^+225^—125. 

9.  Expand  («+2)4.  Ans.     a4+8a3-f-24a2+32a+16 

10.  Expand  (1—  £a)4.  tins.  1—  2a-r-|  --  g+Y6 

11.  Expand  (a-r-6+c)2. 

12.  Expand  (a—2&)3. 

13.  Expand  (1—  2x)5. 


EVOLUTION.  U3 

EVOLUTION. 
CHAPTER  II. 

(Art.  70).  Evolution  is  the  converse  of  involution,  or  the  ex- 
ti  action  of  roots,  and  the  main  principle  is  to  observe  how  powers 
are  formed,  to  be  able  to  trace  the  operations  back.  Thus,  to 
square  «,  we  double  its  exponent,  (Art.  65),  and  make  it  a2. 
Square  this  and  we  have  a4.  Cube  a2  and  we  have  a6.  Take 
the  4th  power  of  x  and  we  have  x4.  The  nth  power  of  x3  is 
xs>\ 

Now,  if  multiplying  exponents  raises  simple  literal  quantities 
to  powers,  dividing  exponents  must  extract  roots.  Thus,  the 

square  root  of  a4  is  a2.  The  cube  root  of  a2  must  be  a*.  The 
cube  root  of  a  must  have  its  exponent,  (1  understood,)  divided  by 

3,  which  will  make  a*' 

Therefore  roots  are  properly  expressed  by  fractional  expo- 
nents, i 

The  square  root  of  a  is  a2,  and  the  exponents,  5,  I,  ^,  &c.  in- 
dicate the  third,  fourth,  and  fifth  roots.  The  6th  root  of  y?  is 

s_ 

a?6;  hence  we  perceive  that  the  numerators  of  the  exponent  in- 
dicate the  power  of  the  quantity,  and  the  denominator  the  root 
of  that  power. 

(Art.  71.)  The  square  of  ax  is  aV2.  We  square  both 
factors,  and  so,  for  any  other  power,  we  raise  all  the  factors  to 
the  required  power.  Conversely,  then,  we  extract  roots  by 
taking  the  required  root  of  all  the  factors.  Thus  the  cube  root 
of  8x*  is  2x. 

A  root  that  can  not  be  exactly  expressed  is  called  a  surd,  or 
irrational  root.  The  square  root  of  2  can  not  be  exactly  ex- 
pressed, hence,  it  is  called  a  surd.  A  root  merely  expressed  by 
die  radical  sign  (J ),  or  by  a  fractional  exponent,  is  called  a 
radical  quantity.  The  cube  root  of  9  may  be  symbolically  ex- 
pressed thus,  ^/9,  or  (9)3,  and  this  is  a  radical  quantity;  but 
numerically  the  root  can  not  be  exactly  expressed ;  it  is,  there- 
fore, a  surd.  Surd  roots  may  be  found  approximately. 
10 


114  ELEMENTS  OP  ALGEBRA. 

The  square  root  of  64a6  is  obviously  8a3,  and  from  this  and 
the  preceding  examples  we  draw  the  following 

RULE.  For  the  extraction  of  the  roots  of  monomials.  Ex- 
tract the  root  of  the  numeral  coefficients  and  divide  the  exponent 
of  each  letter  by  the  index  of  the  root. 

EXAMPLES. 

1.  What  is  the  square  root  of  49a2#4  ?  Ans.     7ax2 

2.  What  is  the  square  root  of  25c1062  1  Jins.     5c5b. 

3.  What  is  the  square  root  of  2Qax  ?  fins. 


In  20,  the  square  factor  4  can  be  taken  out  ;  the  other  factor 
is  5.  The  square  root  of  4  is  2,  which  is  all  the  root  we  can 
take  ;  the  root  of  the  other  factors  can  only  be  indicated  as  in  the 
answer. 


4.  What  is  the  square  root  of  12a2  1  £ns. 

5.  What  is  the  square  root  of  144a2cV/?       Jins.  I2ac*xy. 

6.  What  is  the  square  root  of  36.r4  ?  fins.     db6a?. 

(Art.  72.)  The  square  root  of  algebraic  quantities  may  be 
taken  with  the  double  sign,  as  indicating  either  plus  or  minus, 
for  either  quantity  will  give  the  same  square,  and  we  may  not 
know  which  of  them  produced  the  power.  For  example,  the 
square  root  of  16  may  be  either  -j-4  or  —  4,  for  either  of  them, 
when  multiplied  by  itself,  will  produce  16. 

The  cube  root  of  a  plus  quantity  is  always  plus,  and  the  cube 
root  of  a  minus  quantity  is  always  minus.  For  -\-2a  cubed 
gives  -|-8a3,  and  —  2a  cubed  gives  —  8a3,  and  a  may  represent 

any  quantity  whatever. 

i 

EXAMPLES.  . 

1.  What  is  the  cube  root  of  125a3?  tins.     5a. 

2.  What  is  the  cube  root  of  —  64#6  ?  tfns.     —  4x* 

3.  What  is  the  cube  root  of  —  216«y  ?  rfns.     —  6«y3 

4.  What  is  the  cube  root  of  729  A12?  dns.     9a*x* 

5.  What  is  the  cube  root  of  3205  ?  Jlns.            ~ 


EVOLUTION.  115 

6.  What  is  the  4th  root  of  256«V  ?  £ns. 

7.  What  is  the  4th  root  of  16«? 

8.  What  is  the  4th  root  of  64x*y2  ? 

N.  B.  The  1th  root  is  the  square  root  of  the  square  root. 

9.  What  is  the  4th  root  of  20a#  ?  Ans. 

10.  What  is  the  square  root  of  75  ?  £ns. 

75=25X3. 

4a2#4 

11.  Required  the  square  root  of  -5-2"  •  ^n 

.  32«3#5 
13.  Required  the  square  root  of  —  -  .          <flns. 


N.  B.  Reduce  the  fraction  as  much  as  possible,  and  then  ex- 
tract the  root. 

13.  Required  the  square  root  of  ,    ^    .  *fl.ns.     ±—r* 

128a  4 


14.  Required  the  nth  root  of  -sr—  ..  rfns.     -r-. 

nn 


a3  2.  ~1.  -1 

15.  Required  the  nth  root  of  =-.  Am.     an  b"  cn 

Observing  that  j—  =b~}c~l 


CHAPTER  m. 
To  extract  roots  of  compound  quantities. 

(Art.  72.)  We  shall  commence  this  investigation  by  confining 
our  attention  to  square  root,  and  the  only  principle  to  guide  us  is 
the  law  of  formation  of  squares.  The  square  of  a-{-b  is  «2-}- 
2ab-\-bz.  Now  on  the  supposition  that  we  do  not  know  the  root 
'of  this  quantity  to  be  a-}-b,  we  are  to  find  it  or  extract  it  out  of 
the  square 


We  know  that  a2,  the  first  term,  must  have  been  formed  by  the 
multiplication  of  a  into  itself,  and  the  next  te-m  is  2«X#.    That 


110  ELEMENTS  OF  ALGEBRA. 

is  twice  the  root  of  the  first  term  into  the  second  term  of  the  root, 
Hence  if  we  divide  the  second  term  of  the  square  by  twice  the 
root  of  the  first  term,  we  shall  obtain  b,  the  second  term  of  the 
root,  and  as  b  must  be  multiplied  into  itself  to  form  a  square,  we 
add  b  to  2«,  and  have  2«-f-/;,  which  we  call  a  divisor. 

OPERATION. 


a2 

2a-\-b}2ab+b* 
2ab+b2 

We  take  a  for  the  first  term  of  the  root,  and  subtract  its  square 
(«2)  from  the  whole  square.  We  then  double  a  and  divide  2ab 
by  2a  and  we  find  6,  which  we  place  in  both  the  divisor  and  quo- 
tient. Then  we  multiply  2a-\-b  by  b  and  we  have  2ab-{-bz,  to 
subtract  from  the  two  remaining  terms  of  the  square,  and  in  this 
case  nothing  remains. 

Again,  let  us  take  rt+6-f-c,  and  square  it.  We  shall  find  its 
square  to  be 


a2 


2ab+b2 
2ab+b2 


2aci-2bc+cz 

By  operating  as  before,  we  find  the  first  two  terms  of  the  root 
to  be  a-}-b,  and  a  remainder  of  2ac-\-2bc-{-cz.  Double  the  root 
already  found,  and  we  have  2a-}-2b  for  a  partial  divisor.  Divide 
the  first  term  of  the  remainder  2ac  by  2a,  and  we  have  c  for  the 
third  term  of  the  root,  which  must  be  added  to  2a-\-2b  to  com- 
plete the  divisor.  Multiply  the  divisor  by  the  last  term  of  the 
root  and  set  the  product  under  the  three  terms  last  brought  down, 
and  we  have  no  remainder. 


EVOLUTION.  117 

Again,  let  us  take  a-\-b-\-c  to  square;  but  before  we  square  it 
let  the  single  letter  s=a-}-b. 

Then  we  shall  have  s-f-c  to  square,  which  produces 
s2 -f-2sc+c2.  To  take  the  square  root  of  this  we  repeat  the  first 
operation,  and  thus  the  root  of  any  quantity  can  be  brought  into 
a  binomial  and  the  rule  for  a  binomial  root  will  answer  for  a  root 
containing  any  number  of  terms  by  considering  the  root  already 
found,  however  great,  as  one  term. 

Hence  the  following  rule  to  extract  the  square  root  of  a  com- 
pound quantity. 

Arrange  the  terms  according  to  the  powers  of  some  letter, 
beginning  with  the  highest,  and  set  the  square  root  of  the  first 
term  in  the  quotient. 

Subtract  the  square  of  the  root  thus  found  from  the  first 
term  and  bring  down  the  next  two  terms  for  a  dividend. 

Divide  the  first  term  of  the  dividend  by  double  of  the  root 
already  found,  and  set  the  result  both  in  the  root  and  in  t/ie 
divisor. 

Multiply  the  divisor,  thus  completed,  by  the  term  of  the  root 
last  found,  and  subtract  the  product  from  the  dividend,  and  so  on. 

EXAMPLES. 
1.  What  is  the  square  root  of 


2a2-f26 


2a2-f-46—  2  —  402—  86+4 


2.  What  is  the  square  root  of    1  —  4b-\-4b2-\-2y  —  iby-{-y*1 

Ans.     1  —  2b  -f-y. 

8.  What  is  the  sq  lare  root  of    4z4—  4or3-{-13rJ—  6#-J-9  ? 

Ans.     2x?—  a?-f3. 

4.  What  is  the  sa  lore  root  of    4z4—  lGx3+24x2—  16#-f4  ? 


118  ELEMENTS  OF  ALGEBRA. 

5.  What  is  the  square  root  of   I6x*  +24x?+S9x*+6Qx-{'  100? 

Am. 


«    What  is  the  square  root  of    4x*  —  16#3+8a:2-f-16#-{-4  ? 

Ans. 
7.  What  is  the  square  root  of 

3?+2xy-\-y2-\-6xz  -\-6yz-\-  9z*l  dm. 


8.  What  is  the  square  root  of 

Ans. 

a2  b" 

0.  What  is  the  square  root  of    ^  —  2-\-  -?  ? 


a     b        b     a 

Jlns.    T  -  or  -  T 

b     a        a    b 


1O.  What  is  the  square  root  of    a.*  —  fcB* 


^-T/     or  y-x 
^j^L-  d.TM.cM^..  fW&d,    J0  &  : 

(Art.  73.)  Every  square  root  will  be  equally  a  root  if  we  change 
the  sign  of  all  the  terms.  In  the  first  example,  for  instance,  the 
root  may  be  taken  —  a2  —  26+2,  as  well  as  az-\-2b  —  2,  for  either 
one  of  these  quantities,  by  squaring,  will  produce  the  given 
square.  Also,  observe  that  every  square  consisting  of  three 
terms  only,  has  a  binomial  root. 

(Art.  74.)  Algebraic  squares  may  be  taken  for  formulas,  cor- 
responding to  numeral  squares,  and  their  roots  may  be  extracted 
in  the  same  way,  and  by  the  same  rule. 

For  example,  a-f  b  squared  is  az-}-2ab-{-b2,  and  to  apply  this 
to  numerals,  suppose  a  =40  and  6=7. 

Then  the  square  of  40  is         «2=1600 

2ab=  560 
62=     49 


Therefore,  (47)2=2209 

Now  the  necessary  divisions  of  this  square  number,  2209,  are 
not  visible,  and  the  chief  difficulty  in  discovering  the  root  is  to 
make  these  separations. 


EVOLUTION.  119 

The  first  observation  to  make  is  that  the  square  of  10  is  100, 
of  100  is  10000,  and  so  on.  Hence,  the  square  root  of  any 
square  number  less  than  100  consists  of  one  figure,  and  of  any 
square  number  over  100  and  less  than  10000  of  two  figures,  and 
so  on.  Every  two  places  in  a  power  demanding  one  place  in 
its  root. 

Hence,  to  find  the  number  of  places  or  figures  in  a  root,  we 
must  separate  the  power  into  periods  of  two  figures,  beginning 
at  the  unit's  place.  For  example,  let  us  require  the  square  root 
of  22-09.  Here  are  two  periods  indicating  two  places  in  the 
root,  corresponding  to  tens  and  units.  The  greatest  square  in  22 
is  16,  its  root  is  4,  or  4  tens  =40.  Hence  a=40. 

22  09(40+7^47 
a?=  16  00 


2a-K>=80-{-7=87      )609 
609 

Then  2a=80,  which  we  use  as  a  divisor  for  609,  and  find  it 
is  contained  7  times.  The  7  is  taken  as  the  value  of  b,  and 
2«+6,  the  complete  divisor,  is  87,  which  multiplied  by  7  gives 
the  two  last  terms  of  the  binomial  square.  2a6-|-62=560+49 
—609,  and  the  entire  root  40+7=47  is  found. 

Arithmetically,  a  may  be  taken  as  4  in  place  of  40,  and  1600 
as  16,  the  place  occupied  by  the  16  makes  it  16  hundred,  and 
the  ciphers  are  superfluous.  Also,  2«  may  be  considered  8  in 
place  of  80,  and  8  in  60  (not  in  609)  is  contained  7  times,  <fec. 

If  the  square  consists  of  more  than  two  periods,  treat  it  as  two, 
and  obtain  the  two  superior  figures  of  the  root,  and  when  obtain- 
ed bring  down  another  period  to  the  remainder,  and  consider  the 
root  already  obtained  as  one  quantity,  or  one  figure. 

For  another  example,  let  the  square  root  of  399424  be  ex- 
tracted. 


39-94-24  ||  632 
36 


123 


394 
369 


1262 


2524 
2524 


120  ELEMENTS  OF  ALGEBRA. 

In  this  example,  if  we  disregard  the  local  value  of  the  figures, 
we  have  a=6,  2«=12,  and  12  in  39,  3  times,  which  gives  6=3 
Afterwards  we  suppose  «=63,  and  2a=126,  126  in  252,  2  times, 
or  the  second  value  of  b  =-2.  In  the  same  manner,  we  would 
repeat  the  formula  of  a  binomial  square  as  many  times  as  we 
have  periods. 

EXERCISES    FOR    PRACTICE. 

1.  What  is  the  square  root  of  8836?  Jlns.  94. 

2.  What  is  the  square  root  of  106929?  Jlns.  327. 

3.  What  is  the  square  root  of  4782969?  Jlns.  2187. 

4.  What  is  the  square  root  of  43046721?  Jlns.  6561. 

5.  What  is  the  square  root  of  387420489?  Jlns.  19683. 

When  there  are  whole  numbers  and  decimals,  point  off  periods 
both  ways  from  the  decimal  point,  and  make  the  decimal  places 
even,  by  annexing  ciphers  when  necessary,  extending  the  deci- 
mal as  far  as  desired.  When  there  are  decimals  only,  commence 
pointing  off  from  the  decimal  point. 

EXAMPLES. 

1.  What  is  the  square  root  of  10-4976?  Jlns.  3-24. 

2.  What  is  the  square  root  of  3271-4207?     Jlns.  57-19+. 

3.  What  is  the  square  root  of  4795-25731  ?     JZns.  69-247+. 

4.  What  is  the  square  root  of    -0036  ?  Ans.     -06 

5.  What  is  the  square  root  of    -00032754?     rfns.  -01809+. 

6.  What  is  the  square  root  of    -00103041  ?       rfns.     -0321. 
(Art.  75.)  As  the  square  of  any  quantity  is  the  quantity  mul- 
tiplied by  itself,  and  the  product  of  T  by  j    (Art.  64.)  is    p  ; 

hence  to  take  the  square  root  of  a  fraction  we  must  extract  the 
square  root  of  both  numerator  and  denominator. 

A  fraction  may  be  equal  to  a  square,  and  the  terms,  as  given, 
not  square  numbers  ;  such  may  be  reduced  to  square  numbers. 


EVOLUTION.  Itl 

EXAMPLES. 

What  is  the  square  root  of    yYj  • 

Observe    Tyf  =  gf  .                      Hence  the  square  root  is  f  . 

1.  What  is  the  square  root  of    T9/¥  ?  rfns.  £. 

3.  What  is  the  square  root  of    Jr||  ?  Ans.  J. 

3.  What  is  the  square  root  of    \\\±  ?  Ans.  f  . 

4.  What  is  the  square  root  of     fUJ?  ^w*.  £• 

When  the  given  fractions  cannot  be  reduced  to  square  terms, 
reduce  the  value  to  a  decimal,  and  extract  the  root,  as  in  the  last 
article. 

CHAPTER  IV. 

To  extract  the  cube  root  of  compound  quantities. 

(Art.  76.)  We  may  extract  the  cube  root  in  a  similar  manner 
as  the  square  root,  by  dissecting  or  retracing  the  combination  of 
terms  in  the  formation  of  a  binomial  cube. 

The  cube  of  a+b  is  as-\-3a?b-\-3ab2-\-b*  (Art.  67).  Now 
to  extract  the  root,  it  is  evident  we  must  take  the  root  of  the* 
first  term  (a3),  and  the  next  term  is  3a26.  Three  times  the  square 
of  the  first  Utter  or  terra  of  the  root  multiplied  by  the  2d  term 
of  the  root. 

Therefore  to  find  this  second  term  of  the  root  we  must  divide 
the  second  term  of  the  power  (3a26)  by  three  times  the  square 
of  the  root  already  found  (a). 


When  we  can  decide  the  value  of  6,  we  may  obtain  the  com- 
plete divisor  for  the  remainder  after  the  cube  of  the  first  term  is 
subtracted,  thus  : 

The  remainder  is     3azb-}-3ab2-\-b3 

Take  out  the  factor  6,  and  3a2-f-3a6-j-62   is   the  complete 

divisor  for  the  remainder.     But  this  divisor  contains  6,  the  very 

term  we  wish  to  find  by  means  of  the  divisor  ;  hence  it  must  be 

found  before  the  divisor  can  be  completed.     In  distinct  algebraic 

11 


122  ELEMENTS  OF  ALGEBRA. 

quantities  there  can  be  no  difficulty,  as  the  terms  stand  separate, 
and  we  find  b  by  dividing  simply  3a?b  by  3a2  ;  but  in  numbers 
the  terms  are  mingled  together,  and  b  can  only  be  found  by  trial. 

Again,  the  terms  3a2-^-3ab-\-b2  explain  the  common  arithme- 
tical rule,  as  3a2  stands  in  the  place  of  hundreds,  it  corresponds 
with  the  words  :  "  Multiply  the  square  of  the  quotient  by  300y" 
"and  the  quotient  by  30,"  (3a,)  &c. 

By  inspecting  the  various  powers  of  a-\-b,  (Art.  67,)  we  draw 
the  following  general  rule  for  the  extraction  of  roots  : 

Arrange  the  terms  according  to  the  powers  of  some  letter  ; 
take  the  required  root  of  the  first  term  and  place  it  in  the  quo- 
tient :  subtract  its  corresponding  power  from  the  first  term,  and 
bring  down  the  second  term  for  a  dividend. 

Divide  this  term  by  twice  the  root  already  found  for  the  SQUARE 
root,  three  times  the  square  of  it  for  the  CUBE  root,  four  times 
the  third  power  for  the  fourth  root,  &c.,  and  the  quotient  will 
be  the  next  term  of  the  root.  Involve  the  whole  of  the  root, 
thus  found,  to  its  proper  power,  which  subtract  from  the  given 
quantity,  and  divide  the  first  term  of  the  remainder  by  the  same 
divisor  as  before:  proceed  in  this  manner  till  the  whole  root  i9 
determined. 

EXAMPLES. 
1.  What  is  the  cube  root  of   x«-}-Qx5  —  4Qx?-\-QQx  —  64  ? 


a* 
Divisor  3#4)  6r>=  1st  remainder. 


Divisor  3#4     )   —  I2#4=2d  remainder. 


2.  What  is  the  cube  root  of    27a3+108«2+144a+64  ? 

Jlns.     3«-f-4. 

a-  What  is  the  cube  root  of    a3— Qazx+I2ax9'— 8x*  ? 

Jlns.     a — 2x 


EVOLUTION.  123 

4.  What  is  the  cube  root  of    a,'6— 3^5+5x3— 3^—1  1 

Ans.     xz — x — 1 

5.  What  is  the  cube  root  of    a3 — Gcfb+lZab2 — Qb 3? 

Am.     a — 26. 


O  I 

6.  What  is  the  cube  root  of    x*-\-3x-\ h- 


*  Extract  the  fourth  root  of 

a4+8a3-f-24a2-f-32a-H  6(a-}-2 
a4 


Ans.    x-\ — . 


4a3)       8a3,     &c. 


(Art.  77.)  To  apply  this  general  rule  to  the  extraction  of  the 
cube  root  of  numbers,  we  must  first  observe  that  the  cube  of  10 
is  1000,  of  100  is  1000000,  &c.;  ten  times  the  root  producing 
1000  times  the  power,  or  one  cipher  in  the  root  producing  3  in 
the  power ;  hence  any  cube  within  3  places  of  figures  can  have 
only  one  in  its  root,  any  cube  within  6  places  can  have  only  two 
places  in  its  root,  &c.  Therefore  we  must  divide  off  the  given 
power  into  periods  consisting  of  three  places,  commencing  at  the 
unit.  If  the  power  contains  decimals,  commence  at  the  unit 
place,  and  count  three  places  each  way,  and  the  number  of  pe- 
riods will  indicate  the  number  of  figures  in  the  root.' 

EXAMPLES. 

1.  Required  the  cube  root  of     12812904. 

12-812-904(234 
0=2          «3=  8 


Divisor        3a2=12  )48 


12167  =  (23)3 
3(23)2=   1587)        6459    (4 


12812904  = 


124  ELEMENTS  OF  ALGEBRA. 

Here  12  is  contained  in  48,  4  times;  but  it  must  be  remem- 
bered that  12  is  only  a  trial  or  partial  divisor;  when  completed 
it  will  exceed  12,  and  of  course  the  next  figure  of  the  root  can- 
not exceed  3. 

The  first  figure  in  the  root  was  2.  Then  we  assumed  a~2. 
Afterwards  we  found  the  next  figure  must  be  3.  Then  we  as- 
sumed a=23.  To  have  found  a  succeeding  figure,  had  there 
been  a  remainder,  we  should  have  assumed  a=234,  &c.,  and 
from  it  obtained  a  new  partial  divisor. 

2.  What  is  the  cube  root  of  148877?  Ans.     53. 

3.  What  is  the  cube  root  of  571787?  dns.     83. 

4.  What  is  the  cube  root  of  1367631  ?  rfns.     111. 

5.  What  is  the  cube  root  of  2048383  ?  rfns.     127. 

6.  What  is  the  cube  root  of   16581375?  dm.     255. 

7.  What  is  the  cube  root  of  44361864  ?  Ans.     354. 
§.  What  is  the  cube  root  of  100544625?              Ans.     465. 

(Art.  78.)  When  the  power  is  not  complete,  and,  of  course, 
its  root  surd,  the  methods  of  direct  extraction  are  all  too  tedious 
to  be  much  used,  and  several  eminent  mathematicians  have  given 
more  brief  and  practical  methods  of  approximation. 

One  of  the  most  useful  methods  may  be  investigated  as  follows  : 

Suppose  a  and  a-\-c  two  cube  roots,  c  being  very  small  in 
relation  to  «.  a3  and  «3+3a2c+3«c2+c3  are  the  cubes  of  the 
supposed  roots. 

Now  if  we  double  the  first  cube  (a3),  and  add  it  to  the  second, 
we  shall  have  3a>+3<?c+3ac>+c>. 

• 

If  we  double  the  second  cube  and  add  it  to  the  first,  we  shall 
have 


As  c  is  a  very  small  fraction  compared  to  «,  the  terms  con- 
taining c2  and  c3  are  very  small  in  relation  to  the  others,  and 
the  relation  of  these  two  sums  will  not  be  materially  changed  by 
rejecting  those  terms  containing  c2  and  c3,  and  the  same  will 

lhen  be  3«»+3  etc 

And          3o3+6a'c 


EVOLUTIOIS7  125 

The  ratio  of  these  terms  is  the  same  as  the  ratio*  of  a-\-c  to 
a-f2c. 

Or  the  ratio  is  H ; — 


But  the  ratio  of  the  roots  a  to  «+£>  is  1  -\ — . 

a 

Observing  again  that  c  is  supposed  to  be  very  small  in  rela- 
tion to  a.  the  fractional  parts  of  the  ratios  — ; —  and  -  are  both 

a-f-c  a 

small  and  very  near  in  value  to  each  other.  Hence  we  have 
found  an  operation  on  two  cubes  which  are  near  each  other  in 
magnitude,  that  will  give  a  proportion  very  near  in  proportion  to 
their  roots ;  and  by  knowing  the  root  of  one  of  the  cubes,  by  this 
ratio  we  can  find  the  other. 

For  example,  let  it  be  required  to  find  the  cube  root  of  28,  true 
to  4  or  5  places  of  decimals.  As  we  wish  to  find  the  cube  root 
of  28,  we  may  assume  that  28  is  a  cube.  27  is  a  cube  near  in 
value  to  28,  and  the  root  of  27  we  know  to  be  3. 

Hence  a,  in  our  investigation,  corresponds  to  3  in  this  exam- 
ple, and  c  is  unknown;  but  the  cube  of  a-\-c  is  28,  and  a8 
is  27. 

Then     27  28 

2  2 


54  56 

Add       28  27 


Sums     82       :     83     : :     3     :     a-f-c     very  nearly. 

Or  (a-J-c)=27429=3'03658+,  which  is  the  cube  root  of  28, 
true  to  5  places  of  decimals. 

By  the  laws  of  proportion,  which  we  hope  more  fully  to  in- 
vestigate in  a  subsequent  part  of  this  work,  the  above  propor- 
tion, 82  :  83  : :  a  :  a-}-c,  may  take  this  change . 
82  :  1  : :  a  :  c 

Hence,  c=TV  c  being  a  correction  to  the  known  root, 
which,  being  applied,  will  give  the  unknown  or  sought  root. 

From  what  precedes,  we  may  draw  the  following  rule  for  find- 
ing approximate  cube  roots : 


126  ELEMENTS  OF  ALGEBRA. 

RULE.  Take  the  nearest  rational  cube  to  the  given  number, 
and  call  it  an  assumed  cube ;  or,  assume  a  root  to  the  given 
number  and  cube  it.  Double  the  assumed  cube  and  add  the 
given  number  to  it ;  also,  double  the  given  number  and  add 
the  assumed  cube  to  it.  Then,  by  proportion,  as  the  first  sum 
is  to  the  second,  so  is  the  known  root  to  the  required  root.  Or 
take  the  difference  of  these  sums,  then  say,  as  double  of  the 
assumed  cube,  added  to  the  number,  is  to  this  difference,  so  is 
the  assumed  root  to  a  correction. 

This  correction,  added  to  or  subtracted  from  the  assumed  root, 
as  the  case  may  require,  will  give  the  cube  root  very  nearly. 

By  repeating  the  operation  with  the  root  last  found  as  an  as- 
sumed root,  we  may  obtain  results  to  any  degree  of  exactness ; 
one  operation,  however,  is  generally  sufficient. 

EXAMPLES. 

1.  What  is  the  approximate  cube  root  of  120? 

tins.     4-93242-K 

2.  What  is  the  approximate  cube  root  of  8*5  ? 

£ns.     2-0408-f. 

3.  What  is  the  approximate  cube  root  of  63  ? 

Jlns.     3-97905-}-. 

4.  What  is  the  approximate  cube  root  of  515  ? 

tins.     8-01559-)-. 

5.  What  is  the  approximate  cube  root  of  16? 

The  cube  root  of  8  is  2,  and  of  27  is  3 ;  therefore  the  cube 
root  of  16  is  between  2  and  3.  Suppose  it  2-5.  The  cube  of 
this  root  is  15-625,  which  shows  that  the  cube  root  of  16  is  a 
little  more  than  2-5,  and  by  the  rule 


31-25 

32 

16 

15-625 

47-25 

:      47-625 

:  :     2-5     :     to  the 

required  root. 

47-25 

:           -375 

::     2-5     :     -01984 

Assumed  root 

2-50000 

Correction 

•01984 

Approximate  root          2-51984. 


EVOLUTION.  127 

We  give  the  last  as  an  example  to  be  followed  in  most  cases 
where  the  root  is  about  midway  between  two  integer  numbers. 

This  rule  may  be  used  with  advantage  to  extract  the  root  of 
perfect  cubes,  when  the  powers  are  very  large. 

EXAMPLE. 

The  number  22-069'810'125  is  a  cube;  required  its  root. 

Dividing  this  cube  into  periods,  we  find  that  the  root  must 
contain  4  figures,  and  the  superior  period  is  22,  and  the  cube  root 
of  22  is  near  3,  and  of  course  th<  whole  root  near  3000;  but  less 
than  3000.  Suppose  it  2800  and  cube  this  number.  The 
cube  is  21952000000,  which  being  less  than  the  given  number, 
shows  that  our  assumed  root  is  not  large  enough. 

To  apply  the  rule,  it  will  be  sufficient  to  take  six  superior 
figure^  of  the  given  and  assumed  cubes.  Then  by  the  rule, 

220698 
2 

439040     4413% 
220698     219520 


659738   :   660916   : :   2800 
659738 


659738   :     1178   ::   2800 
2800 


942400 
2356 

659738)3298400(5 
3298690 

Assumed  root,     2800 
Correction,  5 

True  root,  2805 

The  resuit  of  the  last  proportion  is  not  exactly  5,  as  will  be 
seen  by  inspecting  the  work ;  the  slight  imperfection  arises  from 
the  rule  being  approximate,  not  perfect. 

When  we  have  cubes,  however,  we  can  always  decide  the  unit 
figure  by  inspection,  and,  in  the  present  example,  the  unit  figure 


128  ELEMENTS  OF  ALGEBRA. 

in  the  cube  being  5,  the  unit  figure  in  the  root  must  be  5,  as  no 
other  figure  when  cubed  will  give  5  in  the  place  of  units. 

[For  several  other  abbreviations  and  expedients  in  extracting 
cube  root  in  numerals,  see  Robinson's  Arithmetic.] 

(Art.  79.)  To  obtain  the  4th  root,  we  may  extract  the  square  root 
of  the  square  root.  To  obtain  the  6th  root,  we  may  take  the 
square  root  first,  and  then  the  cube  root  of  that  quantity. 

To  extract  odd  roots  of  high  powers  in  numeral  quantities  is 
very  tedious  and  of  no  practical  utility  ;  we  therefore  give  no  ex- 
amples. 

(Art.  80.)  Some  radical  quantities  may  be  simplified,  and  oth- 
ers are  not  susceptible  of  simplification.  For  instance,  the  square 
root  of  75,  written  ^/75,  is  equal  to  5  times  the  square  root  of 
3;  that  is,  ^/75=5^/3.  But  the  square  root  of  71  can  not  be 
fully  expressed  except  by  the  sign,  thus,  ^/71,  because  71  con- 
tains no  square  factor. 

It  is   obvious   that   *J4Q==*J8:r5==*iJS  *^Ss=S(5)lj  but 

39  and  41  containing  no  cube  factor,  their  cube  roots  can  only 
be  indicated  by  the  radical  sign  over  them. 

When  simplification  is  possible,  it  can  be  effected  by  the 
following 

RULE.  Separate  the  quantity  into  two  factors,  one  of  which 
is  a  perfect  power  of  the  required  root.  Extract  the  root  of 
that  factor  and  prefix  the  result  as  a  coefficient  to  the  other 
factor  placed  under  the  radical  sign. 

We  give  the  following  examples  for  practice  : 

1.  Reduce  the  square  root  of  75  to  lower  terms,  or  reduce 

tins,     5^/3. 

2.  Reduce    J98a2  to  lower  terms.  Jlns.     7«>/2. 

3.  Reduce    jVZx9-y  to  lower  terms.  Jlns.     2x,j3y. 


4.  Reduce   3«/54.T4  to  lower  terms.  Jlns.     3x 

5.  Reduce   4  ^/T08    to  lower  terms.  Jlns. 


6.  Reduce    Jx?  —  «V   to  lower  terms.         Jlns.     xjx  —  a2. 


EVOLUTION.  20  129 


7.  Reduce   3«/32a3  to  lower  terms.  Ans.     2«3>/4. 


8.  Reduce    J'ZScfx?  to  lower  terms.  Ans.     2axj7a 

9.  Reduce    74  J  to  lower  terms. 

Where  terms  under  the  radical  are  fractional,  it  is  expedient 
to  reduce  the  denominator  to  a  power  corresponding  to  the  radi- 
cal sign ;  then  by  extracting  the  root  there  will  be  no  fraction 
under  the  radical. 

The  above  example  may  be  treated  thus  : 


-y  K'-J X33=ftV88.       Aw. 

We  divided  ||  into  the  factors  •£•§  and  y  ;  the  first  factor  is  a 
square ;  the  other  factor,  y ,  we  multiply  both  numerator  and 
denominator  by  3,  to  make  the  denominator  a  square. 

In  like  manner  reduce  the  following  : 

10.  Reduce   37^  to  more  simple  terms.       Ans.     %*J\Q. 

11.  Reduce   37V  to  more  simple  terms.        Ans. 

12.  Reduce  to  more  sim^e  terms.         Ans. 


13.  Reduce   3fJa^-\-a3b2  to  more  simple  terms. 

Ans.     a 


14.  Reduce   ,J~  to  more  simple  terms.  Ans. 

(Art.  81.)  Radical  quantities  may  be  put  into  one  sum,  or  the 
difference  of  two  may  be  determined,  provided  the  parts  essen- 
tially radical  are  the  same. 

Thus  the  sum  of  ^/8  and  ,/72  is  8^/2  and  their 
difference  is  4^/2 

For     ^8= 

And      72= 


Sum  872 

Difference   472" 

When  radical  quantities  are  not  and  cannot  be  reduced  to  the 


130  ELEMENTS  OF  ALGEBRA. 

same  quantity  under  the  sign,  their  sum  and  difference  can  only 
tie  taken  by  the  signs  plus  and  minus. 

EXAMPLES. 


1.  Find  the  sum  and  difference  of  tJ\.Qa2x   and    ,j4a2x. 

Ans.     Sum,  Qajx ;  difference,  2atjjs. 

2.  Find  the  sum  and  difference  of  «/128    and    ^72. 

Jlns.     Sum,  14^/2  ;  difference, 

3.  Find  the  sum  and  difference  of  3^/135   and   3N/40. 

Am.     Sum,  5  3,y5  ;  difference, 

4.  Find  the  sum  and  difference  of  V 108   and   9  V 4. 

Jins.     Sum,  123^/4;  difference,  6  3>/4. 

5.  Find  the  sum  and  difference  of   ^/|   and    J'l. 

Ans.     Sum,  f^2  ;  difference,  J^/2U 

6.  Find  the  sum  and  difference  of  3^/56   and   3>/189. 

•fins.     Sum,  53,y7;  difference,3^/?. 

7.  Find  the  sum  and  difference  of  3Jazb  and 


Am.    Sum,    14-3a6;  difference,   12^— 


(Art.  82.)  We  multiply  letters  together  by  writing  them  one 
after  another,  as  abxy.  If  they  are  numeral  quantities,  their 
product  appears  as  a  number  ;  if  two  or  more  of  them  are  nu- 
meral, the  product  of  these  quantities  will  appear  as  a  number. 

This  fundamental  principle  of  multiplication  may  be  applied  to 
the  multiplication  of  surds.  Let  it  be  required  to  multiply 
5^/2  by  3^/7.  Here  suppose  «=5,  6=3,  x=j2,  y=,Jl. 
Then  the  product  of  5^/2T  by  S^/Y  is  abxy  or  15^/2X7= 
15^14.  Hence,  for  the  multiplication  of  quantities  affected  by 
the  same  radical  sign,  we  draw  the  following 

RULE.  Multiply  the  rational  parts  together  for  the  rational 
part  of  the  product,  and  the  radical  parts  together  for  the 
radical  part  of  the  product. 


EVOLUTION.  131 

EXAMPLES. 

1.  Required  the  product  of  5^/5  and  3^/8. 

Product  reduced,     tins. 

2.  Required  the  product  of  4J12  and  3^/27      Am. 

3.  Required  the   product  of   3,/2~  and  2J~8.          .#ns.     24. 

4.  Required  the  product  of  2yl4  and  3  3^/47 

Ans.     12  V7. 

5.  Required  the  product  of  2^5~and  2^/To.         Jlns. 


(Art.  83.)  We  multiply  different  powers  of  the  same  quantity 
together  by  simply  adding  the  exponents  as,  a3  multiplied  by  a3 
is  a8,  and  this  holds  when  the  exponents  are  fractional,  as  a2  into 
a3;  for  the  product  we  simply  add  J  and  -J,  which  make  |  to 
write  over  a  for  the  product  of  a2  into  a?  which  gives  cfi. 

When  the  radical  quantities  are  different,  as  («)2  and  6^,  and 
the  exponents  different,  are  to  be  multiplied  together,  or  one  to  be 
divided  by  another,  we  better  lay  aside  all  rules  and  apply  that 
powerful  engine,  an  equation.  For  illustration: 

Required  the  product  of  a2  into  b^. 

The  product  is  P,  then  we  have 


Raise  both  members  to  the  6th  power,  then, 


Conceiving  the  second  member  to  be  a  single  quantity,  and 
taking  the  6th  root,  will  give  us 


EXAMPLES. 

1.  Required  the  product  of  («+&)'    and   (a-\-b)*. 

Jlns. 

2.  Required  the  product  of  JT~  and  *JT,          rfns.    (75)« 


132  ELEMENTS  OF  ALGEBRA. 

(Art.  84.)  If  we  divide  one  quantity  by  another,  the  product 
of  the  quotient  and  the  divisor  must  equal  the  dividend. 
Divide  the  QJl2  by  ^/2~:  the  quotient  is  Q. 
Then, 

By  squaring, 

Cubing  8  £6=72. 

Whence,  Q*=9         Q3=3 

3.  Required  the  product  of  2J3  and  3(4)^. 


_        _ 
4.  Required  the  product  of  3^/15  and  ^10. 

Am.  6^/225000. 

EXAMPLES. 

1.  Divide  4^/50  by  2</57  Ans.     2JW. 

2.  Divide  6  yiOO   by  3  y~5~.  Am.     2  y20i 

3.  Divide    Jl   by  B4J'7. 

Let  a=7.  Then  the  example  is,  to  divide  the  a2  by  aa,  or 
a¥  by  a¥  ;  quotient  a?.  As  we  always  subtract  the  exponents 
of  like  quantities  to  perform  division  (Art.  17)  ;  therefore  7¥  must 
be  the  required  quotient. 

4.  Divide   6^/54   by  3J2.  Ans. 

5.  Divide   (tfbzcl?Y  by  A  -#rcs. 

6.  Divide    (I6tf—I2a2x)*    by  2«.  ^?is.     (4a—  3xf. 
(Art.  85.)  In  the  course  of  algebraical  investigations,  we  might 

fall  on  the  square  root  of  a  minus  quantity,  as  J  —  a,  ,J  —  1, 
J  —  b,  &c.,  and  it  is  important  that  the  pupil  should  readily 
understand  that  such  quantities  have  no  real  existence  ;  for  no 
quantity,  either  plus  or  minus,  multiplied  by  itself  will  give 
minus  a,  minus  b,  or  minus  any  quantity  whatever  ;  hence  there 
is  no  value  to  J  —  a,  &c.,  and  such  symbols  are  said  to  be  im- 
possible or  imaginary. 


EVOLUTION.  133 

(Art.  86.)  The  square  and  cube  root  of  any  quantity,  as  a, 
being  expressed  by  .Ja  and  3Ja,  and  as  by  involving  the  root 
we  obtain  the  power,  hence  the  square  of  ,Ja,  is  a;  and  the 
cube  of  I/a,  is  a.  Hence,  removing  the  sign  involves  to  the 
corresponding  power. 

EXAMPLES. 


1.  What  is  the  square  of  J3ax  1  Jlns.     Sax. 

2.  What  is  the  cube  of  V%2  ?  J3ns.     6y2. 


3.  What  is  the  square  of  Ja2  —  x2  ?  JLns.     a2  —  x2. 


4.  What  is  the  square  of  /vL_  ?  rfns.     - 

6-f-z  6+0? 


5.  What  is  the  cube  of  *Jl+a*.  rfns.     l+a. 

(Art.  87.)  When  we  have  two  or  more  quantities,  and  the 
radical  sign  not  extending  over  the  whole,  involution  will  not 
remove  the  radical,  but  will  change  it  from  one  term  to  another. 
Thus,  Jx-\-a  the  square  will  be  x-{-2a,Jx-\-a* ;  the  radical 
sign  is  still  present,  but  not  in  the  first  term. 


PURE  EQUATIONS. 
CHAPTER  IV. 

(Art.  88.)  Pure  equations  in  general  are  those  in  which  a  com- 
plete power  of  the  unknown  quantity  is  contained,  and  no  special 
artifice  is  requisite  to  render  the  power  complete. 

The  unknown  quantity  may  appear  in  one  or  in  several  terms ; 
when  it  appears  in  several,  its  exponents  will  be  regular,  descend- 
ing from  a  higher  to  a  lower  value,  or  the  reverse. 

In  such  cases  we  must  reduce  the  equation  by  evolution. 

Like  roots  of  equal  quantities  are  equal.     (Ax.  8.) 

EXAMPLES. 

1.  Given  3x2 — 9=66  to  find  the  value  of  x. 
Solution,  3,r2=^75  #2=25.     Hence,  #=±5. 


134  ELEMENTS  OF  ALGEBRA. 

We  place  the  double  sign  before  5,  as  we  cannot  determine 
\vhether25  was  produced  by  the  square  of  +5  or  of  —  5. 

In  practical  problems,  the  nature  of  the  case  will  commonly 
determine,  but  in  every  abstract  problem  we  must  take  the 
double  sign. 

S5.  Given  xz  —  6#-|-9=a2  to  find  the  value  of  x. 
By  evolution,  x  —  3=±a.     Hence,  a?=3±#,  Jlns. 

3.  Given  x3—  xz-{-ix—^=asb3  to  find  x. 

Take  cube  root  and  x  —  ^—ab;  and  x=ab-\-\,  Jlns. 

4.  Given  a;4  —  2^+1  =  16  to  find  the  value  of  x. 

Jlns.     x=±^  or 


5.  Given  xz  —  4#+4=64   to  find  the  value  of  x. 

Jlns.     #=10,   or  —  6. 

6.  Given  x2y2-}-2xy-}-l=4:x2y2  and  x=2y  to  find  the  values 
of  x  andy.  Jlns.     #==t,/2^  y 


7.  Given  x2-\-y2=I3  and  x2  —  2/2=5  to  find  the  values  of  x 
and  y.  Jlns.  #=±3  y=±2. 

(Art.  89.)  The  unknown  quantity  of  an  equation  is  as  likely 
to  appear  under  a  radical  sign  as  to  be  involved  to  a  power.  In 
such  cases  we  free  the  unknown  quantity  from  radicals  by  involu- 
tion (Art.  86.),  having  previously  transposed  all  the  terms  not 
under  the  radical  to  one  side  of  the  equation,  the  radical  being 
on  the  other. 


9.  Given    ,Jx-{-l=a — 1    to  find  the  value  of  x. 

By  involution,  #-{-l=a2 — 2a-r-l.     Hence,  x—a2 — 2a. 

10.  Given  ,J  1 2 -f- x =3 -\-Jx  to  find  the  value  of  x. 
Square  both  sides,  and  we  have 


Drop  9-}- x  and   3=6^;   or    Jx~=s.  x=%,    Jim. 


11.  Given    Jx — 16=8 — Jx  to  find   x.         Jlns.     #=25. 


PURE  EQUATIONS.  135 

X — ax      J  x 

12.  Given  — 1=  =  — -   to  find  x. 

v#         x 

Multiply  by  ^/^  observing  that  x  divided  by  x  gives  i ;  and 
we  have  x — ax=l, 

Or  (I—  d)x=l, 

Therefore         x=- . 

1 — a 

13.  Given   — .•= = — r— to  find  the  value  of  x. 


By  clearing  of  fractions,  we  have 

tf+34/H-168=;r  4-42^+152. 
Reducing,  16=8^ 

By  division,  2=  Jx,         or         4=x. 

To  call  out  attention  and  cultivate  tact,  we  give  another  solu- 
tion.    Divide  each  numerator  by  its  denominator,  and  we  have 


Drop  unity  from  both  sides,  and  divide  by  8  ;  we  then  have 
3  4 


Clearing  of  fractions,     3tJx-\-l8=4iJx-{-lQ 

Dropping  equals,  2=  ,Jx.     Hence,  x=4. 

14.  Given   ,J x-{- ,J a-\-x=  . =  to  find  x.     Jlns.  x=\ 

V a-\-x 

2az 

15.  Given     x-}-Jaz-\-x2=j          •  to  find  x. 

Jlns.    a?=a,u 


».  Given     a?+a=Va2+a?V62+ic2  to  find  x. 

b*—4a2 

Jlns.    x= — 

40 


136  ELEMENTS  OF  ALGEBRA. 

Jftx — 2      4tjQx — 9 
17.  briven   ^ =    ^    — to  find  a?.*       Jlns. 


18.  Given  V64-f-  x2-^Sx= — to  find  a:,     dns.  # 


15 


19.  Given   J&+x-}-Jx=  —  -  to  find  a?.       rfns.    #=4 

V5+* 

20.  Given    Jx+Jx—Jx  —  Jx=-(  --^—j-  )    to  find  x. 

2\#-rv#/ 


25 
Jin,.    *—. 


Because     a2—  62  =(a 

We  infer  that     5z—  9=(^/5a?-f3)  (j5x—3) 

Therefore,  =^5^  —  3.     The  given  equation   then 

-- 


becomes    .^53:  —  3=1-1—^-  --  . 
2 

Now  assume 


Then  y=l+iy.     Consequently,  2/=2.     Returning  to  equa 
tion  (w5),  we  have  ^5a;  —  3=2 

tj5x=5.         Therefore,         x=5,     Ans. 

Jax—b     3j~ax—2b  Qb* 

22.  Given  ^-—  —  ==-^±  -  to  find  x.       Am.  x=  — 
Jax+b     3,Jax-{-5b 

(Compare  22  with  examples  13,  and  17.) 


23.  Given   (1+^7^+12)  =  l+a?  to  find  x.     rfns. 
21.  Given    ^!!±_1  +  ^=9,  to  find  x. 


*  See  2d  solution  to  Equation  13. 


PURE  EQUATIONS.  137 

25.  Given   a2 — 2ax-{-xz=b,  to  find  x.          £ns.  #=« — Jb 

\  1/3 

26.  Given =-^r    to  find  x. 

Jlns.     x=±i. 

4 

27.  Given  • r— =|,   to  find  x.  Ans.     x=5. 

x2 — 2#-{-l 


28.  Given  -==l-\-.—  to  find  x.         Aw.  x= 


Q/« 

29.  Given       x--x—  9=  to  find  a?. 


30    Given  =-      to  find  a?.  ^ns.    a=4 


31.  Given  —=L  to  find  the  value  of  a?. 


Multiply  the  first  member,  numerator  and  denominator,  by 
x-\-  ,Jx  —  a),  then  both  members  by  a,  and  extract  square  root. 


32.  Given  'ql^  =&>   to  find  a-. 

Assume   a-{-#=2/.     Then  the  equation  becomes 

'=&.     Hence,  v= 


And 


(Art.  90.)  To  resolve  the  following  examples,  requires  a  de- 

gree of  tact  not  to  be  learned  from  rules.     Quickness  of  percep- 

tion is  requisite,  as  well  as  sound  reasoning.     Quickness  to  per- 

ceive the  form  of  binomial  squares,  and  binomial  cubes,  and  a 

12 


138  ELEMENTS  OF  ALGEBRA. 

readiness  to  resolve  quantities  into  simple  or  compound  factors, 
as  the  case  may  require. 

1.  Given    x*-}-2x=9-+-—  to  find  the  value  of  x 

Multiply  by  x,  and  x3+2x*=9x+18. 
Separate  into  factors,  thus  :  (a?-f-2)a?2=(#-{-2)9. 
Divide  by  the  common  factor,  tf-J-2,  and  x*=9,  or  #=±3. 


Given     i  a?2,^~24  £   to  find  the  values  of  x  and  y. 
Add  the  two  equations  together,  and  we  have 


Extract  square  root,  and  x-\-y=±6.  (.#) 

From  the  first  equation  we  have  (x-\-y)x=l.2.    (B) 
Divide  equation  (B}  by  (.#),  and  a?=±2. 

This  example  required  perception  to  recognise  the  binomial 
square,  and  also  to  separate  into  factors. 

|    Q  /» 

3.  Given   xz-{-yz=  -   and   xv=  -   to  find  the  values 
x—y  x—y 

of  x  and  y. 

From  the  first  equation  subtract  twice  the  second,  and 


Therefore,     (x — #)3=1,     and    x — y=l. 
Continuing  the  operation,  we  shall  find  #=3,  and  y=2. 

4.  Given   x2y-{-xy2=lSO  and  xs +y*=  189,  to  find  the  values 
of  x  and  y.  J2ns.     x=5  or  4  ;  y=4  or  5. 

To  resolve  this  problem,  requires  the  formation  of  a  cube,  or 
to  resolve  quantities  into  factors. 

5.  Given    x*-\-y*=(x-{-y}xy,   and    x-}-y=4,   to    find    the 
values  of  x  and  y.  Jlns.     x=2]  2/=2. 

6.  Given  x-{-y  :  x  ::  7  :  5,   and   xy+y*=I26i   to   find  tha 
values  of  x  and  y.  Jlns.     #=±15,  ?/=±6 


PUKE   EQUATIONS.  139 

7.  Given   x—  y  :  y  ::  4  :  5,   and   x2+4y*=l8],   to  find   the 
values  of  x  and  y.  rfns.     #=±9,   y=±5. 


8.  Given    Jx+Jy  >  Jx  —  Jy  ::  4  :  1,    and   x  —  */=16,   to 
find  the  values  of  x  and  y.  Ans.     #=25,   i/=9. 

9    Given  -^+|-f7=9»   to  find  a?.  Am      x=7±. 

y     o    4 

10.  Given  x+y  :  x-y  ::  3  =  1  ?  to  find  x  and 

And         a?3—  y  =56          5 


11.  Given    y?y-}-xyz=30  } 

1       i        5  I  to  find  x  and 

And         +'= 


Observe  that  xy(x-\-y)=x?y-{-xyz. 

Clear  the  2d  equation  of  fractions,  and  y-}-x  or  x-\-y=  —  ~. 

Now  assume  x-\-y=s,  and  xy—p.      Then  the  original  equa- 
tions become  s/?=30 

And  6s=5p 

Equations  which  readily  give  s  and  j9,  and  from  them  we  de- 
termine x  and  y. 

N.  B.  When  two  unknown  quantities,  as  x  and  y,  produce 
equations  in  the  form  of 

x+y=s  (1) 

And  xy=p  (2) 

such  equation  can  be  resolved  in  the  following  manner  : 

Square  (1),  and         x*-}-2xy-{-y2=s2 
Subtract  4  times  (2)         4xy        ==4p 

I  Diff.  is  x2—  2xy+y2=s2—  4p 

By  evolution  x  —  y—Js*  —  4p  (3) 

Add  equation  (1)  and  (3),  and  we  have, 

(4) 


Sub.  (3)  from  (1),  and    2y=s—Js2—4p       (5) 


140  ELEMENTS  OF  ALGEBRA. 

To  verify  equations  (4)  and  (5),  add  them  and  divide  by  2, 
and  we  have  x-\-y=s.  Multiply  (4)  by  (5),  and  divide  by  4, 
and  we  have  xy=p. 

(Art.  91.)  No  person  can  become  very  skilful  in  algebraic 
operations  as  long  as  he  feels  averse  to  substitution;  for  judi- 
cious substitution  stands  in  the  same  relation  to  common  algebra, 
as  algebra  stands  to  arithmetic.  The  last  example  is  an  illustra- 
tion of  this  remark.  To  acquire  the  habit  of  substituting,  may 
require  some  extra  attention  at  first,  but  the  power  and  advantage 
gained  will  a  thousand  fold  repay  for  all  additional  exertion. 

When  two  equations  involve  quantities  in  the  form  (x-\-y)  and 
xy,  or  x—y,  and  xy,  and  any  product  of  these  quantities,  the 
equations  are  easily  solved. 

The  following  examples  will  illustrate  : 


12.  Given   x++y=  197        find      and 
And  x2-\-xy+y2  =133$ 


Put    x  -\-y  =5,       and 
Then      s+p=  19  (£) 

And        s2—  /=133  (E) 

Divide  (B)  by  (.#),  and  we  have  s—p=7,  &c. 

Ans.    #=9  or  4,    y=4  or  9. 

13.    Given      x4-}-2xzy2+y4=l2QQ-^4xy(x2-{-xy-{-y2)      and 
x  —  y=4,    to  find  x  and  y. 

Put  x*-\-y2=s,     and     xy=p 

Then  the  first  equation  becomes     s2=1296  —  4p(s-\-p) 
Multiply  and  transpose,  and     s2-\-4sp-\-4p2=l29Q 

Square  root  »+2p=±36 

But     s-t-2/?=a:2-{-2^-f-2/2==±36 

Therefore  x+=  ±6,         or 


Rejecting  imaginary  quantities,  we  find   x=5   or   —  1,   and 
?/=l    or   —  5. 

X2—'UZ 

14.  Given  --  —=G,  and  x-{-y-\-xy=U,  to  find  the  values 
x  —  y 

of  x  and  y.  Jlns.    x=5  or  1,  y=l  or  5 


PURE  EQUATIONS.  HI 

15.  Given   x*-{-ys=2xy(x-\-y],    and    xy=l6,    to   find   the 
values  of  x  and  y.  Jlns.     #=2^54-2,   7/^2^/5  —  2. 

16.  Given    x3-\-y3—  «,  and   xzy-}-xy*=a,  to  find  the  relative 
values  of  x  and  y.  tfns.     x—y. 

1  7.  Given   x-\-y  :  x  ::  5  :  3,   and   xy=Q,  to  find  a?  and  y. 

Ans.     #=±3,  2/=±2. 

18.  Given    #-f-?/  :  a:  ::  7  :  5,    and  a?7/-l-?/2=126,   to  find  a: 
and  y.  Jlns.     a?  =±15,   i/=±6. 

19.  Given    ,T2-)-?/2=«,   and    xy=b,   to   find   the   values   of 
x  and  y.  Ans.     x=±.*tJa-}-2b-{-%lJa—<2,b. 


(Art.  A)*  Equations  in  the  form  of  a?4  —  2ax2-}-az=b,  require 
for  their  complete  solution,  the  square  root  of  an  expression  in 
the  form  of  a±Jb  ;  for  by  extracting  the  square  root  of  the 
equation,  we  have 


Hence  x=\/  a 

The  right  hand  member  of  this  equation  is  an  expression  well 
known  among  mathematicians  as 

A    BINOMIAL    SURD. 

Expressions  in  this  form  may  or  may  not  be  complete  powers  ; 
and  it  is  very  advantageous  to  extract  the  root  of  such  as  are 
complete,  for  the  roots  will  be  smaller,  and  more  simple  quantities, 
in  the  form  of  a'i^/6',  or  of  .Ja'^jW. 

Let  us  now  investigate  a  method  of  extracting  these  roots  ;  and, 
for  the  sake  of  simplicity,  let  us  square  34-^7. 

By  the  rule-  of  squaring  a  binomial,  we  have     94-6^/7-1-7, 
Or,  le+G/T; 

Conversely,  then,  the  square  root  of  164-6^/7,   is  3+77. 

*  That  the  same  Articles  may  number  the  same  in  both  the  School  and  Col- 
lege  Edition,  we  shall  designate  all  additional  Articles,  in  this  volume,  by 
A,  B,  &c. 


142  ELEMENTS  OF  ALGEBRA. 

But  when  a  root  consists  of  two  parts,  its  square  consists  of  tht 
sum  of  the  squares  of  the  two  parts,  and  twice  the  product  of 
the  two  parts. 

Now  we  readily  perceive  that  16  is  the  sum  of  the  squares  of 
the  two  parts  expressing  the  root;  and  6^/7  ,  the  part  containing 
the  radical,  is  twice  the  product  of  the  two  parts. 

To  find  what  this  root  must  be,  let  a?  represent  one  part  of  the 
root,  and  y  the  other  : 

Then  aHy=16,  (1) 

And  2xy=&J7,  (2) 

Add  equations  (1)  and  (2),  and  extract  square  root,  and  we 

have 


(3) 
Subtract  equation  (2)  from  (1),  and  extract  square  root,  and 


we  have  x—y=^lQ—  Gj7~.        (4) 

Multiply  (3)  and  (4),  and  we  have 

a?2—  2/2=^/256—  252=>/T=:2  ;  (5) 
Add  (1)  and  (5),  and  we  have 


#=,     or 
Sub.  (5)  from  (1),  and  2#2=14,     or    y=, 

Wber»ie  x-\-y,  or  the  square  root  of  16-|-6^7   is  S-f-^/7. 
f  e  shall  now  be  more  general. 

Take  two  roots,  one  in  the  form  of 
and  one  in  the  form  of 
Square  both,  and  we  shall  have  «2±2a>/6-f-6, 

And  a±2,Jab-\-b. 

In  numerals,  and,  in  short,  in  all  cases,  the  sum  of  the  squares 
of  the  two  parts  of  the  root,  as  («2+6),  in  the  first  square,  and 
(a-f-6),  in  the  second,  contains  noradical  sign  ;  and  the  sum  of 
these  rational  parts  may  be  represented  by  c  and  c',  and  the 
squares  represented  in  the  form  of 


PURE   EQUATIONS.  143 

c  ±2a«/6, 
or  of         c'±2jab. 

Hence,  generally,  if  we  represent  the  parts  of  the  roots  by  y 
and  y,  we  shall  have  x?-}-yZ:=  the  sum  of  the  rational  parts,  and 
2xy=  the  term  containing  the  radical. 

The  signs  to  x  and  y  must  correspond  to  the  sign  between 
the  terms  in  the  power.  If  that  sign  is  minus,  one  of  the  signs 
of  the  root  will  be  minus ;  it  is  indifferent  which  one. 

EXAMPLES. 

1.  What  is  the  square  root  of   ll-j-6^/2"?         Jlns. 

2.  What  is  the  square  root  of  7-f-4A/3~7          Jlns. 

3.  What  is  the  square  root  of  7 — 2^/10? 

Jlns.     75— J2   or    </2— J&1 

4.  What  is  the  square  root  of  94+42^5'?       Jlns.  7+3^5". 

5.  What  is  the  square  root  of  28+1073?       Jlns.    5-f-^ 

6.  What  is  the  square  root  of 

np-}-2mz — 2mtJnp-\-m2 1 

In  this  example  put  a—np-{-m*,  and  x  and  y  to  represent 
the  two  parts  of  the  root, 

Then  x2+y2—m2+a, 

and  2xy  = — 2m  Ja. 

Jlns.     db(  Jnp-{-mz — m) . 


.  What  is  the  square  root  of  bc-+-2bjbc  —  b21 

Jlns. 


8.  What  is  the  sum  of 

Jlns.     10. 


9.  What  is  the  sum  of  ^/l  1+6^/2   and  A/7— 


3+^/5*. 
10 


Am.    S-\-2^5  or  4. 


144  ELEMENTS  OF  ALGEBRA. 

In  a  similar  manner  we  may  extract  the  cube  root  of  a  bino- 
mial surd,  when  the  expression  is  a  cube  ;  but  the  general  solu- 
tion involves  the  solution  of  a  cubic  equation,  and,  of  course, 
must  be  omitted  at  this  place  ?  and,  being  of  little  practical  utility, 
we  may  omit  it  altogether. 

(Art.  92.)  Fractional  exponents  are  at  first  very  troublesome 
to  young  algebraists  ;  but  such  exponents  can  always  be  ban- 
ished from  pure  equations  by  substitution.  For  the  exponents 
of  all  such  equations  must  be  multiples  of  each  other;  otherwise 
they  would  not  be  pure,  but  complex  equations. 

To  make  the  proper  substitution,  place  the  unknown  quanti- 
ties having  the  lowest  exponents  equal  to  other  simple  quantities, 
as  P  and  Q  ;  and  let  this  be  a  general  direction. 

EXAMPLES. 

£        1  4.          2 

1.  Given   x3-[-y5  =  G,   and   re3  -}-;ys=20,   to  find  the  values 

of  x  and  y 

2_  \_ 

By  the  above  direction,  put  x3=P,  and  y5  —  Q. 
Squaring  these  auxiliaries,  or  assumed  equations, 

And        x^=P2,      and      y~*  =  Q\ 
Now  the  original  equations  become 
=Q         (1) 
=2Q         (2) 

By  squaring  equation  (1),  P2-f2P#-f-Q2=36. 
Subtracting  equation  (2),  we  have  2PQ=IQ. 
Subtracting  this  last  from  equation  (2),  and  we  have 


By  extracting  square  root  P  —  Q=±2 
But  by  equation  (1),          PJrQ=     6 

Therefore,     P=4  or  2,  and  Q=2  or  4, 

fi 
JL&VUVW,         ~  =4  or  2,  and?/5  =2  or  4. 

Square  root  ^=2  or  (2)*  ) 

'3   V  y=32  or   1024 

Cubing  gives  x=S  or  (2)*  J 


PURE  EQUATIONS.  145 

2.  Given  xyz+y=2l,  and  a?y+2/2=333,  to  find  the  values 
of  x  and  y. 

By  comparing  exponents  in  the  two  equations,  we  perceive 
that  if  we  put  xy2=P,  and  y=Q,  the  equations  become 

P+Q=  21 
P*+  $2=333 

Solved  as  the  preceding,  gives  P=18,    Q=3. 
From  which  we  obtain     a?=2,   or   T^¥,  y=3  or  18. 


3.  Given    xz+x?y*=  208 

24  to  find  the  values  of  x  and  y. 

And 


2  £ 

Assume         x*=P,     and    y*  =  Q. 

By  squaring  and  cubing  these  assumed  auxiliary  equations, 
we  have  4 


Seek  the  common  measure  (if  there  be  one)  between  208  and 
1053. 

From  the  above  substitution,  the  given  equations  become 
P*-\-P2Q=  208=13.16         (1) 
$3-}-$2p=:1053=13.81         (2) 
Separate  the  left  hand  members  into  factors,  and 
\3.lG  (3) 

13.81  (4) 

Divide  equation  (4)  by  (3),  and  we  have 

Q2     81  &    9 

-^2=—.     Extracting  square  root    -^=- 

9P 
Or  Q——T--     Substitute  this  value  of  Q  in  equation  (1), 

q/33 

And         /M-f-^-=13.16 
4 

Or  4P4-9P2=13.64 
That  is,  13P3=13.6i 
Hence,  P3=64  or  P=4 

13 


146  ELEMENTS  OF  ALGEBRA 

But     «2=/w=64.         Therefore,     #=±8 

As    Q=—      and     P=4,         Q—9     and    #=±27. 

3  3     3  3_  1 

4.  Given  #2-f-#4i/4-H/2  =  1009     =a 

33  r  to  find  a?  and  y. 

And    orM-a^v'-H/3  =582193=6 


Put     x*=P    and    #*  =  # 

Then  ar*=/»    and    y^=(f 
And    a-3  =  /*    and     yz  =  #* 
Our  equations  then  become 

/»_!_  PQ-L.Q* 1  Equations  having  no  fractional  exponents, 

p\\piru\Qi t  f      an(l  are  of  the  same  form  as  in  Problem 

J       12.     (Art.  9 1.) 

Jim.     #=81  or  16,     t/=16  or  8]. 

5.  Given   x-{-x*y*  =  l2  } 

j    j  r  to  find  the  values  of  x  and  y. 

And  y+x*y*=  4  J 

^ns.     x=9,    y=l 

6.  Given  x-l-*V=«    | 

j    j  r  to  find  the  values  of  x  and  y. 

And  y\-x*y*=zb    } 

a  a*  b* 

rt  jyns.    x— — r^-    v= — ;-, 

a+b    9     a-\-b 

333  1 

T.  Given   x2+x*y*=a  [ 

3          j-  to  find  the  values  of  x  and  y 

And  1  3 


8.  Given    Jx-\-Jy  :  Jx  —  Jy  ::  4  :  I,   and   a:—  1/=16,   to 
find  the  values  of  x  and  y.  As.     a?=25,    y=9. 


9.   Given   *+y~  5       tQ  find 

And   ar  =13 

fins.     x=9  or  4,    y=4  or  9 


PURE  EQUATIONS.  147 

JO.  Given   x-\-y  :  x  —  y  ::  3  :  1  }     to    find    the    values  of 
And          *3—  */3=56  $         *  and  y. 

fins.     #=4,    2/=2. 
11.  Given  x  -\-y  =35  } 

r  to  find  the  values  of  x  and  y. 
And   £3+7,3=5    J 


CHAPTER  V. 
Problems  producing  Pure  Equations. 

(Art.  93.)  We  again  caution  the  pupil,  to  be  very  careful  not  to 
involve  factors,  but  keep  them  separate  as  long  as  possible,  for 
greater  simplicity  and  brevity.  The  solution  of  one  or  two  of 
the  following  problems  will  illustrate. 

1.  It  is  required  to  divide  the  number  of  14  into  two  such  parts, 
that  the  quotient  of  the  greater  divided  by  the  less,  may  be  to  the 
quotient  of  the  less  divided  by  the  greater,  as  16  :  9. 

•fins.     The  parts  are  8  and  6. 

Let    x=   the  greater  part.     Then    14  —  x=   the  less. 

x         14  —  x 

Per  question,   ~-  -  :  --  :  :  16:9. 
14  —  x        x 

AT    I   '    1  9X  16(14—  X) 

Multiply  extremes  and  means,  and    —  -  -  =  —  -  -  '- 

14  —  x  x 

Clearing  of  fractions,  we  have  9#2=16(14  —  xf 
By  evolution,  3^=4(14  —  x)  =4.  14  —  4x 

By  transposition,      7#=4.14 
By  division,  #=4.2=8,  the  greater  part. 

14—  "X 

Had  we  actually  multiplied  -         -   by  16,  in   place  of  indi 

rating  it,  the  exact  value  and  form  of  the  factors  would  have  been 
lost  to  view,  and  the  solution  mighthave  run  into  an  adfected  qua- 
dratic equation. 

The  same  remark  may  be  applied  to  many  other  problems, 
and  many  are  put  under  the  head  of  quadratics  that  may  be  re 
duced  by  pure  equations. 


148  ELEMENTS  OF  ALGEBRA. 

2.  Find  two  numbers,  whose  difference,  multiplied  by  the 
difference  of  their  squares,  is  32,  and  whose  sum,  multiplied  by 
the  sum  of  their  squares,  gives  272. 

If  we  put  x=  the  greater,  and  y=  the  less,  we  shall  have 

(z-.yX^-yH  32      (i) 

And        (*+2/)(*2-f3/2)=272         (2) 

Multiply  these  factors  together,  as  indicated,  and  add  the  equa- 
tions together,  and  divide  by  2,  and  we  shall  have 

o;3+y3=152         (3) 

If  we  take  (1)  from  (2),  after  the  factors  are  multiplied,  we 
shall  have  2tri/2-f  2arty=240,  or  xy(x+y}  =  \'2Q  (4) 

Three  times  equation  (4)  added  to  equation  (3)  will  give  a 
cube,  &c.  A  better  solution  is  as  follows  : 

Let  x-}-y=  the  greater  number,  and  x  —  y—  the  less. 
Then  2x=  their  sum,  and  2y=  their  difference. 
Also,    4xy  is  equal    the    difference    of   their   squares,   and 
2x2-\-2yz=  the  sum  of  their  squares. 

By  the  conditions,  2yX4xy=  32 

And  2#(2ori4-2*/2)=272 

By  reduction,        xy*=  4 
And 


By  subtraction,      x3      =64     or     #=4 

Hence,  y=l,  and  the  numbers  are  5  and  3. 

We  give  these  two   methods  of  solution  to  show  how  much 
depends  on  skill  in  taking  first  assumptions. 

3.  From  two  towns,  396  miles  asunder,  two  persons,  Ji  and  Z?, 
set  out  at  the  same  time,  and  met  each  other,  after  traveling  as 
many  days  as  are  equal  to  the  difference  of  miles  they  traveled 
per  day,  when  it  appeared  that  A  had  traveled  216  miles.  How 
many  miles  did  each  travel  per  day?  Let  x=JPs  rate,  nnd  i/= 
fl's  rate. 

Then    x  —  y=  the  days  they  traveled  before  meeting. 

By  question,   (x  —  y)x=21G,   and   (x  —  y)y=lSQ. 

216     180         6     5 

vonsequently,      --  =  -  or  -=-. 
y'       x        y         x    y 


PURE   EQUATIONS.  14 

Therefore,     ?/=|x,  which  substitute  in  the  first  equation,  and 

x2 
eve  have   (z—  f#)#=216,   or   --=216=6X6X6. 

By  evolution,   #=36;    therefore   y=  30. 

4.  Two  travelers,  A  and  B,  set  out  to  meet  each  other,  A 
leaving  the  town  C,  at  the  same  time  that  B  left  D.  They  traveled 
the  direct  road  between  C  and  D  ;  and  on  meeting,  it  appeared 
that  A  had  traveled  18  miles  more  than  B,  and  that  A  could  have 
gone  £'s  distance  in  1  5|  days,  but  B  would  have  been  28  days 
in  going  ^'s  distance.  Required  the  distance  between  C 
and  D. 

Let  x=  the  number  of  miles  Ji  traveled. 
Then  x  —  18=  the  number  B  traveled. 

x  —  18 

=M  s  daily  progress. 


154 


—  =J9's  daily  progress. 
Sffo 


Therefore,     ,  :  ,-18  :  :  f 

if      4(x— 


Divide  the  denominators  by  7,  and  extract  square  root,  and  we 


Therefore,     #=72  ;  and  the  distance  between  the  two  towns 
is  126  miles. 

5.  The  difference  of  two  numbers  is  4,  and  their  sum  multi- 
plied  by   the  difference  of  their  second   powers,  gives  1600. 
What  are  the  numbers  1  Jlns.  12  and  8. 

6.  What   two  numbers  are  those  whose  difference  is  to  the 
less   as  4  to  3,    and   their   product,  multiplied  by  the  less,  is 
equal  to  504?  dns.  Hand  6. 


150  ELEMENTS  OF  ALGEBRA. 

7.  A  man  purchased  a  field,  whose  length  was  to  its  breadth 
as  8  to   5.     The  number  of  dollars    paid  per  acre  was   equal 
to  the   number   of   rods   in  the   length  of  the  field;    and   the 
number  of  dollars  given  for  the  whole  was  equal  to  13  times 
the  number  of  rods  round   the  field.     Required  the  length  and 
breadth  of  the  field. 

Jlns.  Length  104  rods,  breadth  65  rods. 
Put  8,r=the  length  of  the  field. 

8.  There  is  a  stack  of  hay,  whose  length  is  to  its  breadth  as 
5  to  4,  and  whose  height  is  to  its  breadth  as  7  to  8.      It  is  worth 
as  many  cents  per  cubic  foot  as  it  is  feet  in  breadth ;   and  the 
whole  is  worth  at  that  rate  224  times  as  many  cents  as  there  are 
square  feet  on  the  bottom.  Required  the  dimensions  of  the  stack. 

Put  5x  =  the  length. 

Jlns.  Length  20  feet,  breadth  16  feet;  height  14  feet. 

9.  There  is  a  number,  to  which  if  you  add  7,  and  extract  the 
square  root  of  the  sum,  and  to  which  if  you  add  16  and  extract 
the  square  root  of  the  sum,  the  sum  of  the  two  roots  will  be  9. 
What  is  the  number  ?  Jlns.  9. 

Put   x2 — 7=    the   number. 

10.  Jl  and  B  carried  100  eggs  between  them  to  market,  and 
each  received  the  same    sum.      If  A  had  carried  as  many  as 
BI  he  would  have  received  18  pence  for  them;  and  if  B  had 
taken  as  many  as  .#,  he  would  have  received  8  pence.     How 

many  had  each  ?  Jlns.  Jl  40,  and  B  60. 

/ 

11.  The  sum  of  two  numbers  is  6,  and  the  sum  of  their  cubes 
is  72.     What  are  the  numbers  ?  Jlns.  4  and  2. 

12.  One  number  is  a2  times  as  much  as  another,  and  the  pro- 
duct of  the  two  is  b2.     What  are  the  numbers  ? 

b 

Jlns. —  and  ab. 
a 

13.  The   sum   of  two   numbers   is  100,  the   difference    of 
their  square  roots  is  2.     What  are  the  numbers? 

Jlns.  36  and  64. 


PURE  EQUATIONS.  151 

Put  x=  the  square  root  of  the  greater  number, 
And  y=  the  square  root  of  the  less  number  ;  or 
Put  x-\-y=  the  square  root  of  the  greater,  &c. 

14.  It  is  required  to  divide  the  number  18  into  two  such  parts, 
that  the  squares  of  those  parts  may  be  to  each  other  as  25  to  16 

Let  x=  the  greater  part.     Then  18  —  x=  the  less. 
By  the  condition  proposed,  x*:  (18  —  #)2::25:  16. 
Therefore,         16^=25(18—  a?)2 
By  evolution,    4a?=±  5(18  —  a:) 

If  we  take  the  plus  sign,  as  we  must  do  by  the  strict  enuncia- 
tion of  the  problem,  we  find  #=10.  Then  18  —  #=8. 

And  (10)2:  (8)2::25:  16 
If  we  take  the  minus  sign,  we  shall  find  a?=90. 

Then  18—  x=  18—  90=—  72. 

And  (90)2:  (  —  72)2::25:  16;  a  true  proportion,  correspond- 
ing to  the  enunciation  ;  but  18  in  this  case  is  not  the  number 
divided,  it  is  the  difference  between  two  numbers  whose  squares 
are  in  proportion  of  25  to  16. 

15.  It  is  required   to  divide  the  number  a  into  two  such 
parts  that  the  squares  of  those  parts  may  be  in  proportion  of  b 
to  c. 

Let  a?=  one  part,  then  a  —  x=  the  other. 
By  the  condition,         x2  :  (a  —  x)2  ::b:c 
Therefore,  cx2=b(a  —  x)2 

By  evolution,  tjcx=±ljb(a  —  x) 

Taking  the  plus  sign,       x=-~—r-    and    a  —  X- 


ajb  —  ajc 

Faking  the  minus  sign,    x=-rr^  —  -r-    and    a  —  a?=   .,    *—-. 

Jb  —  ,Jc 


Prob.  14,  is  a  particular  case  of  this  general  problem,  in  which 
a=  18,  6=25,  and  c=16;  and  substituting  these  values  in  the  re- 
sult, we  find  a?=10,  and  #=90,  as  before. 

If  we  take  b=c,  the  two  divisions  will  be  equal,  each  equal 
to  5  a,  when  the  plus  sign  is  used  ;  but  when  the  minus  sign  is 


152  ELEMENTS  OF  ALGEBRA. 

ajb        ajb 
used,   :r=—  —  —  r=i~^~t   a  symbol  of  infinity,  as  the  denomi- 

A/^  -  V  ^ 

nator  is  contained  in  the  numerator  an  infinite  number  of  times. 
(Art.  58.)     The  other  part,  a—  x 

symbol  of  infinity;  and  the  two  parts, 
ajb  ajc        fl 


Jb-Jc       (Jb-Jc) 


It  may  appear  absurd,  that  the  two  parts,  both  infinite  and 
having  a  ratio  of  equality,  (which  they  must  have,  if  b=C)  can 
fetill  have  a  difference  of  a.  But  this  apparent  absurdity  will 
vanish,  when  we  consider  that  the  two  parts  being  infinite  in 
comparison  to  our  standards  of  measure,  can  have  a  difference 
of  any  finite  quantity  which  may  be  great,  compared  with 
small  standards  of  measure,  but  becomes  nothing  in  comparison 
with  infinite  quantities.  See  (Art.  60.) 

Application  of  the  foregoing  Problem. 

(Art.  94.)  It  is  a  well  established  principle  in  physics  that 
light  and  gravity  emanating  from  any  body,  diminish  in  inten- 
sity as  the  square  of  the  distance  increases. 

Two  bodies  at  a  distance  from  each  other,  and  attracting  at  a 
given  point,  their  intensities  of  attraction  will  be  to  each  other 
as  the  masses  of  the  bodies  directly  and  the  squares  of  their 
distances  inversely.  Two  lights,  at  a  distance  from  each  other, 
illuminating  at  a  given  point,  will  illuminate  in  proportion  to  the 
magnitudes  of  the  lights  directly,  and  the  squares  of  their  dis- 
tances inversely. 

These  principles  being  admitted  — 

16.  Whereabouts  on  the  line  between  the  earth  and  the  moon 
will  these  two  bodies  attract  equally,  admitting  the  mass  of  the 
earth  to  be  75  times  that  of  the  moon,  and  their  distance  asunder 
30  diameters  of  the  earth? 

Represent  the  mass  of  the  moon  by  c, 

and  the  mass  of  the  earth  by  6, 

their  distance  asunder  by  a. 


PURE  EQUATIONS.  153 

The  distance  of  the  required  point  from  the  earth's  centre, 
represent  by  x.  Then  the  remaining  distance  will  be  (a  —  x). 

Now  by  the  principle  above  cited,  x2  :  (a  —  xf  ::b:c. 

This  proportion  is  the  same  as  appears  in  the  preceding  gen- 
eral problem  ;  except  that  we  have  here  actually  made  the  ap- 
plication, and  must  give  the  definite  values  to  a,  b  and  c. 

ajb  ajc 

As  before,    x  =—  rrn  —  j-  and  a  —  x=—y  — 


fl=30,         6=75,         c=l. 
x=      ^      =26&,  nearly.     Hence,  a  —  a?=3.1,  nearly. 

If  we  take  the  second  values  for  the  two  distances,  from  the 

ajb 

neral  result,  namely,  #=-—  v    .      a 
V^  —  vc 

give  the  numeral  values,  we  shall  have 


ajb  —  ajc 

general  result,  namely,  #=-—  v    .      and   a  —  x=   .     v  .  ,  and 
V^  —  vc  v°  —  vc 


«—  *=—  3.9,  nearly. 

These  values  show  that  in  a  line  beyond  the  moon,  at  a 
distance  of  3.9  the  diameters  of  the  earth,  a  body  would  be 
attracted  as  much  by  the  earth  as  by  the  moon,  and  the  value 
of  (a  —  x)  being  minus,  shows  that  the  distance  is  now  counted 
the  other  way  from  the  moon,  not  as  in  the  first  case  towards 
the  earth  ;  and  the  real  distance,  30,  corresponding  to  a  in  the 
general  problem,  is  now  a  difference. 

We  may  make  very  many  inquiries  concerning  the  intensity 
of  attraction  on  this  line,  on  the  same  general  principle. 

For  example,  we  may  inquire,  whereabouts,  on  the  line  be- 
tween the  earth  and  moon  will  the  attraction  of  the  earth  be  1  6 
times  the  attraction  of  the  moon? 

Let  x=  the  distance  from  the  earth. 
Then  a  —  x—  the  distance  from  the  moon. 

The  attraction  of  the  earth  will  be  represented  by  —  • 


154  ELEMENTS  OF  ALGEBRA. 

0 


The  attraction  of  the  moon  at  the  same  point  will  be 
By  the 


(«-*)' 
b        16c 

3H=? 

T>          i    •  »Jb        4VC 

By  evolution,  ^_=-t--^— 

x         a — x 
Clearing  of  fractions,     ajb — ljbx:=4tjcx. 

Using  the  plus  sign,    x=—  ^     .  =20.5,  nearly. 
Using  the  minus  sign,  x—  ,,     A    ,   —  55.7,    nearly,    or 


25.7  diameters  of  the  earth  beyond  the  moon. 

Observe  that  the  4  which  stands  as  a  factor  to  Jc  is  the 
square  root  of  16,  the  number  of  times  the  intensity  of  the 
earth's  attraction  was  to  exceed  that  of  the  moon. 

If  we  propose  any  other  number  in  the  place  of  16,  its 
square  root  will  appear  as  a  factor  to  «/c;  we  may  therefore 
inquire  at  what  distance  the  intensity  of  the  earth's  attraction 
will  be  n  times  that  of  the  moon,  and  the  answer  will  be  from 
the  earth  in  a  line  through  the  moon, 

and  aJb 

Jb — .Jnc 

The  same  application  that  we  have  made  of  this  general  prob- 
lem to  the  two  bodies,  the  earth  and  the  moon,  may  be  made 
to  any  two  bodies  in  the  solar  system ;  and  the  same  application 
we  have  made  to  attraction  may  be  made  to  light,  whenever 
we  can  decide  the  relative  intensity  of  any  two  lights  at  any 
assumed  unity  of  distance. 

(Art.  95.)  This  problem  may  be  varied  in  its  application  to 
meet  cases  where  the  distances  are  given,  and  the  compara- 
tive intensities  of  light  or  attraction  are  required. 

For  example,  the  planet  Mars  and  the  moon  both  transmit 
the  sun's  light  to  the  earth  by  reflection,  and  we  now  inquire 


PURE  EQUATIONS.  155 

the  relative  intensities  of  their  lights  at  given  distances,  and 
in  given  positions. 

If  the  surface  of  Mars  and  that  of  the  moon  were  equal, 
they  would  receive  the  same  light  from  the  sun  at  equal  distance 
from  that  luminary  ;  but  at  different  distances  equal  surfaces 
•vould  receive  light  reciprocally  proportional  to  the  squares  of 
their  distances. 

The  surfaces  of  globular  bodies  are  in  proportion  to  the  squares 
of  their  diameters.  Now  let  M  represent  the  diameter  of  Mars 
and  m  the  diameter  of  the  moon.  Also,  let  R  represent  the  dis- 
tance of  Mars  from  the  sun,  and  r  the  distance  of  the  moon  from 
the  sun. 

Then  the  quantity  of  light  received  by  Mars  may  -be  expressed 

M2 

by    -^j  ;    and   the   relative    quantity   received    by   the    moon 

m2 
by    —  .     But  these  lights,  when  reflected  to  the  earth,  must  be 

diminished  by  the  squares  of  the  distances  of  these  two  bodies 
from  the  earth.  Now  if  we  put  D  to  represent  the  distance  of 
Mars  from  the  earth,  and  d  the  distance  of  the  moon,  we  shall 

M2 
have    ™~n2  for  the  relative  illumination  by  Mars  when  the  whole 


m 


enlightened  face  of  that  planet  is  towards  the  earth,  and  ~^-=   for 

the  light  of  the  full  moon. 

When  the  whole  illuminated  side  of  Mars  is  turned  towards 
the  earth,  which  is  the  case  under  consideration,  (if  we  take  the 
whole  diameter  of  the  body,)  it  is  then  in  opposition  to  the  sun, 
and  gives  us  light,  we  know  not  how  much,  as  we  have  no 
standard  of  measure  for  it  ;  but  we  can  make  a  comparative  mea- 
sure of  one  by  the  other,  and  therefore  the  light  of  Mars  in  this 
position  may  be  taken  as  unity,  and  in  comparison  with  this  let 
us  call  the  light  of  the  full  moon  x. 

M2         m2 


Theretor*  *- 


156  ELEMENTS  OF  ALGEBRA. 

As  the  value  of  a  fraction  depends  only  on  the  relation  of  the 
numerator  to  the  denominator,  to  find  the  numeral  value  of  x,  it 
will  be  sufficient  to  seek  the  relation  of  m  to  M>  of  R  to  r,  and 
of  D  to  d. 

7)/=4000  miles  nearly,  and  m=2150  ;  hence,  -r>=^- 

Jfl     80 

/?=H4000000,     and     r=95000000  ;      or       —  =^ 

r      95 


/)=144000000—  95000000=49000000   or   ; 

a       24 


That  is,  in  round  numbers,  the  light  of  the  full  moon  is  twenty- 
seven  thousand  six  hundred  times  the  light  of  Mars,  when  that 
planet  is  brightest,  in  its  opposition  to  the  sun. 

We  will  add  one  more  example  by  the  way  of  farther  illustra- 
tion. 

What  comparative  amount  of  solar  light  is  reflected  to  the 
earth  by  Jupiter  and  Saturn,  when  those  planets  are  in  opposi- 
tion to  the  sun  ;  —  the  relative  diameter  of  Jupiter  being  to  that 
of  Saturn  as  111  to  83,  and  the  relative  distances  of  the  Earth, 
Jupiter  and  Saturn,  from  the  sun,  being  as  10,  52  and  95,  re- 
spectively ? 

.#iw.  Taking  the  light  reflected  by  Saturn  for  unity,  that  by 
Jupiter  will  be  expressed  by  24.T\\  nearly. 

The  philosophical  student  will  readily  perceive  a  more  ex- 
tended application  of  these  principles  to  computing  the  relative 
light  reflected  to  us  by  the  different  planets  ;  but  we  have  gone  to 
the  utmost  limit  of  propriety,  in  an  elementary  work  like  this. 

From  Art.  94th  to  the  end  of  this  chapter  can  hardly  be  said 
to  be  algebra  ;  it  is  natural  philosophy,  in  which  the  science  of 
algebra  is  used  ;  however,  we  would  offer  no  apology  for  thus 
giving  a  glimpse  of  the  utility,  the  cui  bono,  and  the  application 
of  algebraic  science. 


QUADRATIC  EQUATION.  157 

SECTION  IV. 

QUADRATIC  EQUATIONS. 
CHAPTER  I. 

(Art.  96.)  Quadratic  equations  are  either  simple  or  compound. 
A  simple  quadratic  is  that  which  involves  the  square  of  the  un- 
known quantity  only,  as  or*=£;  which  is  one  form  of  pure  equa- 
tions, such  as  have  been  exhibited  in  the  preceding  chapter. 

Compound  quadratics,  or,  as  most  authors  designate  them,  ad- 
fectf.d  quadratics,  contain  both  the  square  and  the  first  power 
of  the  unknown  quantity,  and  of  course  cannot  be  resolved  as 
simple  equations. 

All  compound  quadratic  equations,  when  properly  reduced, 
may  fall  under  one  of  the  four  following  forms  : 

(1)  x*+2ax=b 

(2)  x2—  2ax=b 

(3)  Xs—  2ox=—  -b 

(4)  x«-f  2ax=—b 

If  we  take  z-f-a  and  square  the  sum,  we  shall  have 


If  we  take  x  —  a  and  square,  we  shall  have 


If  we  reject  the  3rd  terms  of  these  squares,  we  hare 
3~-\-2ax,  and  x*  —  2ax 

The  same  expressions  that  we  find  in  the  first  members  of  the 
four  preceding  theoretical  equations. 

It  is  therefore  obvious  that  by  adding  a*  to  both  sides  of  the 
preceding  equations,  the  first  members  become  complete  squares. 
But  in  numeral  quantities  how  shall  we  find  the  quantity  corres- 
ponding to  cr  ?  We  may  obtain  a*  by  the  formal  process  of 
taking  half  the  coefficient  of  the  first  power  of  x,  or  the  half  of 
2«  or  —  2«,  which  is  a  or  —  a,  the  square  of  either  being  a*. 

Hence,  when  any  equation  appears  in  the  form  of  x2±2«x= 
rt&  ,we  may  render  the  first  member  a  complete  square,  and  effect 
a  solution  by  the  following 


158  ELEMENTS  OF  ALGEBRA. 

RULE.  Add  the  square  of  half  the  coefficient  of  the  lowest 
power  of  the  unknown  quantity  to  the  first  member  to  complete 
its  square;  add  the  same  to  the  second  member  to  preserve  the 
equality. 

Then  extract  the  square  root  of  both  members,  and  ive  shall 
have  equations  in  the  form  of 


Transposing  the  known  quantity  a  and  the  solution  is  accom- 
plished. 

In  this  manner  we  find  the  values  of  x  in  the  four  preceding 
equations,  as  follows : 

(i)  *=. 

(2)  *= 

(3)  *= 

(4)  *=. 


When  b  is  greater  than  a2  equations  (3)  and  (4)  require  the 
square  root  of  a  negative  quantity,  and  there  being  no  roots  to 
negative  quantities,  the  values  of  x  in  such  cases  are  said  to  be 
imaginary. 

The  double  sign  is  given  to  the  root,  as  both  plus  and  minus 
will  give  the  same  power,  and  this  gives  rise  to  two  values 
of  the  unknown  quantity  ;  either  of  which  substituted  in  the 
original  equation  will  verify  it. 

After  we  reduce  an  equation  to  one  of  the  preceding  forms, 
the  solution  is  only  substituting  particular  values  for  a  and  b  ; 
but  in  many  cases  it  is  more  easy  to  resolve  the  equation  as  an 
original  one,  than  to  refer  and  substitute  from  the  formula. 

(Art.  97.)  We  may  meet  with  many  quadratic  equations  that 
would  be  very  inconvenient  to  reduce  to  the  form  of  x2-{-2ax=b; 
for  when  reduced  to  that  form  2a  and  b  may  both  .be 
troublesome  fractions. 

Such  equations   may  be  left  in  the  form  of 


An  equation  in  which  the  known  quantities,  c,  6,  and  c,  are  all 
whole  numbers,  and  at  least  a  and  b  prime  to  each  other. 


QUADRATIC  EQUATIONS.  159 

We  now  desire  to  find  some  method  of  making  the  first  mem- 
ber of  this  equation  a  square,  without  making  fractions.  We 
therefore  cannot  divide  by  0,  because  b  is  not  divisible  by  a, 
the  two  letters  being  prime  to  each  other  by  hypothesis.  But 
the  first  term  of  a  binomial  square  is  always  a  square.  There- 
fore, if  we  desire  the  first  member  of  our  equation  to  be  convert- 
ed into  a  binomial  square,  we  must  render  the  first  term  a 
square,  and  we  can  accomplish  that  by  multiplying  every  term 
by  a. 

The  equation  then  becomes 

a2x2-}-bax=ca 
Put  y=ax.     Then  yz+by=ca 

Complete  the  square  by  the  preceding  rule,  and  we  have 

.b2          .b2 
y  4-%-f  j=«H-j« 

We  are  sure  the  first  member  is  a  square  ;  but  one  of  the  terms 
is  fractional,  a  condition  we  wished  to  avoid ;  but  the  denomina- 
tor of  the  fraction  is  4,  a  square,  and  a  square  multiplied  by  a 
square  produces  a  square. 

Therefore,  multiply  by  4,  and  we  have  the  equation 

4y2+4by+b2=4ca-{-b2 

An  equation  in  which  the  first  member  is  a  binomial  square  and 
not  fractional. 

If  we  return  the  values  of  y  and  yz  this  last  equation  becomes 
4a2x2+4abx-\-b2=4ac-}-b2 

Compare  this  with  the  primitive  equation 
axz-}-bx= c. 

We  multiplied  this  equation  first  by  0,  then  by  4,  and  in  ad- 
dition to  this  we  find  b2  on  both  sides  of  the  rectified  equation,  b 
being  the  coefficient  of  the  first  power  of  the  unknown  quantity. 
From  this  it  is  obvious  that  to  convert  the  expression  ax2-\-bx 
into  a  binomial  square,  we  may  use  the  following 

RULE  2.  Multiply  by  four  times  the  coefficient  of  x2,  and 
add  the  square  of  the  coefficient  of  x. 

To  preserve  equality,  both  sides  of  an  equation  must  be  mill- 


160  ELEMENTS  OF  ALGEBRA. 

ti  plied  by  the  same  factors,  and  the  same  additions  to  both  sides. 
We  operate  on  the  first  member  of  an  adfected  equation  to 
make  it  a  square,  we  operate  on  the  second  member  to  preserve 
equality. 

(Art.  98.)  For  the  following  method  of  avoiding  fractions  in 
completing  the  square,  the  author  is  indebted  to  the  late  Professor 
T.  J.  Matthews,  of  Ohio. 

Resume  the  general  equation  ax2-\-bx=c 

u  ,     u*  bit 

Assume     x=-          Ihen  aar=  —    and   bx=  — 

a  a  a 

The  general  equation  becomes      —  +  —  =c 

a      a 

Or  u2+bu=ac 

Now  when  b  is  even,  we  can  complete  the  square  by  the  first 
rule  without  making  a  fraction.  In  such  cases  this  transforma- 
tion is  very  advantageous. 

When  b  is  not  even,  multiply  the  general  equation  by  2,  and 
the  coefficient  of  x  becomes  even,  and  we  have 

2c        (1) 


Assume     #=—  -         Then  2ax2=—   and   2bx=—  — 
2a  2a  2a 

With  these  terms,  equation  (1)  becomes 

u2     2bu 
2a+W 

Or        u*-\-2bu=4ac 
Complete  the  square  by  the  first  rule,  and  we  have 


An  equation  essentially  the  same  as  that  obtained  by  completing 
the  square  by  the  rule  under  (Art.  97.)  ;  for  we  perceive  the  sec- 
ond member  is  the  same  as  would  result  from  that  rule  ;  hence 
this  method  has  no  superior  advantage  except  when  b  is  even,  in 
the  first  instance. 

(Art.  99.)  The  foregoing  rules  are  all  that  are  usually  given 
for  the  resolution  of  quadratic  equations  ;  but  there  are  some 


QUADRATIC  EQUATIONS.  161 

intricate  cases  in   practice  that   we  may  meet  with,  where 
neither  of  the  preceding  rules  appear  practical  or  convenient. 
To  master  these  with  skill  and  dexterity,  we  must  return  to  a 
more  general  and  comprehensive  knowledge  of  binomial  squares, 
x* -\-2ax-\- a2  is  a  simple  and  complete  binomial  square.     Let 
us  strictly  examine  it,  and  we  shall  perceive, 
1st.  That  it  consists  of  three  terms ; 

2d.  Two  of  its  terms,  i\\e  first  and  the  third,  are  squares; 
3d.  The    middle    term    is    twice    the  product  of  the  square 
roots  of  the  first  and  last  term. 

Now  let  us  suppose  the  third  term,  a\  to  be  lost,  and  we  have 
only  xz-\-2ax.  We  know  these  two  terms  cannot  make  a  square, 
as  a  binomial  square  must  consist  of  three  terms.* 

We  know  also  that  the  last  term  must  be  a  square. 
Let  it  be  represented  by  t*. 

Then,  by  hypothesis,  xz^-2ax-^t*  is  a  complete  binomial 
square. 

It  being  so,  2xt—2ax,  by  the  third  observation  above. 

Therefore,  t=a  and  t*=a2 
Thus   az  is  brought  back. 

!•  Again,  4a2-}-4ab  are  the  first  and  second  terms  of  a  bi- 
nomial square ;  what  is  the  3rd  term  ? 

Let   t2  represent  the  third  term. 
Then      4«2-f  4ab-}-t2  is  a  binomial  square. 
Hence,  4at—4ab  or  t=b  and  /2=62 

That  is,  t2  represented  the  3d  term,  and  b2  is  the  identical  3d 
lerm,  and  4a2  -\-4ab-\-b2  is  the  actual  binomial  square  whose  root 
Is  2a+b. 

2.  36?/2-}-36?/  are  the  first  and  2d  terms  of  a  binomial  square, 
what  is  the  3d  term  ?  JLns.  9. 

3.  -\ \-9  are  the  2d  and  3d  terms  of  a  square,  what  is  the 

x  ! 

first?  Ans.    —? 

x2- 

*  In  binomial  surds  two  terras  may  make  a  square,  and  this  may  condemn 
the  technicality  here  assumed ;  but  it  is  nothing  against  the  spirit  of  this  ar- 
ticle. 

14 


162  ELEMENTS  OF  ALGEBRA. 

4Qx2 

4.  --  49   are  the  1st  and  2d  terms  of  a  binomial  square, 

49 

what  is  the  3d  1  Jlns.     —  . 

tt 

5.  9?/2  —  Qy   are  the  1st  and  2d  terms  of  a  binomial  square, 
what  is  the  3d?  rfns.     1. 

6.  tfx*-\-bx  are  the  1st  and  2d  terms  of  a  binomial  square, 

b2 


what  is  the  3d  'I  Jlns. 

4a 
( 

7.  8  la;2  —3   are  the  1st  and  3d  terms  of  a  binomial  square, 
what  is  the  2d  or  middle  term  ?  fins.     ±18. 


8.  2/2  —  Sx^y   are  the  1st  and  2d  terms  of  a  binomial  square, 
what  is  the  3d  ?  rfns.     I6x. 


9.  --  ?r+36   are  the  2d  and  3d  terms  of  a  binomial  square, 

1  «7 

or2 

what  is  the  1st  ?  Ans.     -—  -. 

36  1 

772 

10.  ^--}-36   are  the  1st  and  3d  terms  of  a  binomial  square, 
301 

what  is  the  middle  term  ?  rfns. 

iy 

11.  If  #-}---   are  the  2d  and  3d  terms  of  a  binomial  square, 

lo 

what  is  the  1st  term  ?  rfns.    4x*. 

Qx* 

12.  The  1st  term  of  a  binomial  square  is—    the  2d  term  is 

u 

•±:12,  what  is  the  3d  term  ?  Ans.    -^-. 


(Art.  100.)  Adfected  quadratic  equations,  after  being  reduced 
to  the  form  of  x?-\-2ax—b,  can  be  resolved  without  any  formality 
of  completing  the  square,  by  the  following  substitution  : 


QUADRATIC  EQUATIONS.  163 

Assume        x—y  —  o, 
Then  a^=?/2—  2a?/+«2 

And          2ax=    -\-2ay-2a* 
By  addition,  x*+2ax=yz  —  az—b 

Hence,       y=±lJb-\-az 

And      x=  —  a-±ilJb-^-a2j  the    same  result  as 
may  be  found  in  equation  (1),  (Art.  96.) 

RULE  FOR  SUBSTITUTION.  Assume  the  value  of  the  unknown 
quantity  equal  to  another  unknown,  annexed  to  half  the  coeffi- 
cient of  the  inferior  power  with  a  contrary  sign. 

(Art.  101.)  For  further  illustration  of  the  nature  of  quadratic 
equations,  we  shall  work  and  discuss  the  following  equation  : 

Given  ^+4^=60,  to  find  x. 

Completing  the  square,  (Rule  1st.)  £2+4aH-4=64. 

Extracting  square  root,  a?-f-2=rb8. 
Hence,  x=6  or  x=  —  10. 

That  is,  either  plus  6,  or  minus  10,  substituted  for  x  in  the 
given  equation,  will  verify  it. 

For  6H-4X  6=60.     Also,  (—  10)2—  4X  10=60 

If  x=6         then         x—  6=0 

Ifa:=—  10   then        z+10=0 

Multiply  these  equations  together,  and  we  have 
x  —  6 
x  +10 


Product,  #M-4aN—  60=0 

Transpose,  and  0^+42?=  60,  the  original  equation. 

Thus  we  perceive,  that  a  quadratic  equation  maybe  considered 
as  the  product  of  two  simple  equations,  and  these  values  of  x  in 
the  simple  equations  are  said  to  be  roots  of  the  quadratic,  and  this 
view  of  the  subject  gives  the  rationale  of  the  unknown  quantity 
having  two  values. 


164  ELEMENTS  OF  ALGEBRA. 

In  equations  where  but  one  value  can  be  found,  we  infer  that 
the  other  value  is  the  same,  and  the  two  roots  equal,  or  one  of 
them  a  cipher. 

EXAMPLES    FOR    PRACTICE. 

1.  Given  x2— 6#— 7=33,  to  find  x.  Jlns.     #=10  or  — 4. 

2.  Given  x2 — 20#= — 96,  to  find  x.  Jlns.     12  or  8. 

3.  Given  x*-{-6x-{-l=92,  to  find  x.  Jlns.     7  or  -—13. 
•    4.  Given  y2+12y=589,  to  find  y.  Jlns.     19  or — 31. 

5.  Given  t/2 — 6y-|- 10=65,  to  find  y.  Jlns.     1 1  or  —5. 

6.  Given  a?-}-12;c-}-2=110,  to  find  x.  Jlns.     6  or  — 18. 

7.  Given  x2—  14z=51,  to  find  x.  Jlns.     17  or  —3. 

8.  Given  a?2-f6a:-|-6=9,  to  find  x.  Jlns.      —  3±2,y3. 

9.  Given  ar4-8a?=12,to  findtf.  Jlns..     — 

10.  Given  a?-{-l2x=lQ,  to  find  x.  Jlns. 

The  reader  will  observe  that  the  preceding  examples  are  in, 
or  can  be  immediately  reduced  to  the  form  of  #2±2a#=6,  and  of 
course  their  solution  is  comparatively  easy.  The  following  are 
mostly  in  the  form  of  ax*-}-bx=c. 

11.  Given  5#*-j-4:e=204,  to  find  x. 

According  to  (Art.  98,)  put  x—-.  Then    5#2=—    and 

4w  w2     4w 

4x— — ,   and  the  equation  becomes   _-f— ^-=204. 
o  o       5 

Clearing  of  fractions,  w2+4tt=1020. 

Completing  the  square  and  extracting  the  root,  we  have, 
w-f-2=rb32,         or         w=30     or     — 34 

But        x=~.      Therefore,    x—  6    or    — — ,     Jlns. 
5  5 

12.  Given  5^+4#=273,  to  find  x.  Am.     7  or  — 74S 

13.  Given  7a?2 — 20z=32,  to  find  x.  Jlns.     4  or  — f . 

14.  Given  25#2 — 20#= —  3,  to  find  x.         Jlns.     |  or       ^. 

15.  Given  21  x2 — 292#= — 500,  to  find  x. 

ins.     Ilifor2 


QUADRATIC  EQUATIONS.  165 

16.  Given  2x2— 5#=117,  to  find  x. 

Here,  as  5  or  b  of  the  general  equation  is  not  even,  we  must 
multiply  the  whole  equation  by  2,  to  apply  the  above  principle ; 
or  we  may  take  Rule  2.  (Art.  97.) 

Multiply  by  8,  and  add  52  or  25  to  both  members. 

Then        16^—40a:4-25=961 

Square  root,    4x — 5=±31.     Hence,  x =9  or — 65. 

(Art.  102.)  It  should  be  observed  that  all  quadratic  equations 
can  be  reduced  to  the  form  of  xz±.2ax=bt  or,  as  most  authors 
give  it,  xz±px=q;  but  when  the  terms  would  become  fractional 
by  such  reduction,  we  prefer  the  form  ax*±bx=±c,  for  the  sake 
of  practical  convenience,  as  mentioned  in  (Art.  97.) 

(Art.  103.)  It  is  not  essential  that  the  unknown  quantity 
should  be  involved  literally  to  its  first  and  second  powers  ;  it  is 
only  essential  that  one  index  should  be  double  that  of  the  other. 
In  such  cases  the  equations  can  be  resolved  as  quadratics.  For 
example,  a?6 — 4^=621  is  an  impure  equation  of  the  sixth 
degree,  yet  with  a  view  to  its  solution,  it  may  be  called  a  quad- 
ratic. For  we  can  assume  y=x3;  then  y2=x6,  and  the  equa- 
tion becomes  yz — 4i/=621,  a  quadratic  in  relation  to  y,  giving 
y=27,  or  —23. 

Therefore,    #"=27  or — 23 


And  a?=3  or  V— 23. 

There  are  other  values  of  x;  but  it  would  be  improper  to  seek 
for  them  now;  such  inquiries  belong  to  the  higher  order  of 
equations. 

3 

For  another  example,  take  x3 — x2  =56,  to  find  the  values 
of  x. 

Here  we  perceive  one  exponent  of  x  is  double  that  of  the  other, 
it  is  therefore  essentially  a  quadratic. 

Such  cases  can  be  made  clear  by  assuming  the  lowest  power 
of  the  unknown  quantity  equal  to  any  simple  letter.  In  the 

3 

present  case  assume  y=x2;  then  y2=#3,   and  the  equation  is 


166  ELEMENTS  OF  ALGEBRA. 


By  Rule  3,  4y*—4y+  1=225 
By  evolution,  2y  —  1=±15 
Hence,  y=&  °r  —  7 

3  3 

And  by  returning  to  the  assumption  y=x^   we    find  #2—  8, 
or  a?4  =2.     Hence,  a?=4;  or,  by  taking  the  minus  value  of  y, 


(Art.  104.)  When  a  compound  quantity  appears  under  differ- 
ent powers  or  fractional  exponents,  one  exponent  being  double 
that  of  the  other,  we  may  put  the  quantity  equal  to  a  single  letter, 
and  make  its  quadratic  form  apparent  and  simple.  For  example, 
suppose  the  values  of  x  were  required  in  the  equation 


Assume     J2 
Then  by  involution,     2#2-}-3#-f  9=?/2         (A] 
And  the  equation  becomes  y2  —  5y=6  (J9) 

Which  equation  gives  y=&  or  —  1.     These  values  of  y,  sub- 
stituted fory  in  the  equation  (fi    give     2,r2-j-3x-j-9=36 

Or 
From  the  first  of  tl^e  we  find  a?=3  or  — 


From  the  las*  we  find  a?=|(  —  3db«y  —  55,)  imaginary  quan- 
tities. 

EXAMPLES. 

I.  Given  (#+12)*+  (a>H2)*  =  6  to  find  the  values  of  x. 

rfns.     x=4  or  6U 

nd  tlie  values   of  or 
Ans.     x—b*—a  or  81  b'1  —  a.* 

*  It  is  proper  to  remark,  that  in  many  instances  it  would  be  difficult  to  verify 
the  equation  by  taking  the  second  values  of  x,  as  by  squaring,  the  minus  quan- 
tity becomes  plus,  and  in  returning  the  values,  there  is  no  method  but  trial  to 
decide  whether  we  shall  take  a  plus  or  a  minus  root.  Hence,  these  second 
answers  are  sometimes  called  roots  of  solution.  In  many  instances  hereafter, 
we  shall  give  the  rational  and  positive  root  only. 


Given  (aH-</)~*-f  2&(#-f-a)*  =3b2,  to  find  the  values   of  ar 


QUADRATIC  EQUATIONa  107 

3.  Given  9;r+4+2,/9a'+4=15,  to  find  the  values  of  x. 

Jlns.     a?=f      J. 

4.  Given   (10+z)*—  (10+ar)*=2,  to  find  x 

Jlns.    a?=6. 

5.  Given   (#—  5)3—  3(tf—  5)2=40,  to  find  x. 

Ans.    o?=9. 

6.  Given  2(1  -far—  a?2)—  (l-f  a;—  ^)^+^=0,  to  find  a% 


7.  Given  s+16—  3(tf+16f=10,  to  find  a;.      Am.  x=Q. 

8.  Given   So.-2'1—  2o?n=8,  to  find  a*.  «#ns.    o?=%/2. 

9.  Given   3^+^=756,   to  find  x.  Jlns.     a?=243. 

Q  -I     rt 

1O.   Given     -  --  r^1"!"    --      to  ^n(^  ^     «^n5-  a7=3  or  1- 


11.  Given   4a?+=39,   to  find  x.  Am.     a?=729. 

12.  Given   ^—2^+6(^—2^+5)^  =  11,  to  find  x. 

Jlns.    x=l 

it2      12ic 

13.  Given  —  -  —  T^==  —  32,  to  find  the  value  of  #.. 

rfns.     a?=152  or  76. 

If  much  difficulty  is  found  in  resolving  this  13th  example,  the 
pupil  can  observe  the  9th  example,  (Art.  99). 

14.  Given   8  la^+  17+^=99,  to  find  the  values  of  x.  ] 

Jlns.     x=l,  or  —  1,  or      |. 

Observe  that  the  1st  and  3d  terms   of  the  first  number  are 
squares,  see  (Art.  99.) 

1        R41      2^2 

15.  Given   81^+17+-,-=-^-+—  +15,  to  find  x.  J 

X  XT  X 

rfns.     x=2  or  —  1J. 

16.  Given   25#2+6+—  ——'-—-,  to  find  the  values  of  x.~- 

Jlns.     a?—  2,  or  —  2,  or       — 

15 


168  ELEMENTS  OF  ALGEBRA. 

IT.  Given   -^--j--|=6§,  to  find  the  values  of  x. 

Ans.     x=7  or  —  11J. 

(Art.  105.)  Equations  of  the  third,  fourth,  and  higher  degrees, 
can  be  resolved  as  quadratics,  provided  we  can  find  a  compound 
quantity  in  the  given  equation  involved  to  its  first  and  second 
power,  with  known  coefficients. 

To  determine  in  any  particular  case,  whether  such  a  com- 
pound quantity  is  involved  in  the  equation,  we  must  transpose 
all  the  terms  to  the  first  member,  and  if  the  highest  power  of 
the  unknown  quantity  is  not  even,  multiply  every  term  of  the 
equation  by  the  unknown  letter  to  make  it  even,  and  then  extract 
the  square  root,  to  two  or  three  terms,  as  the  case  may  require  ; 
and  if  we  find  a  remainder  to  be  any  multiple  or  any  aliquot 
part  of  the  terms  of  the  root,  a  reduction  to  the  quadratic  form 
is  effected ;  otherwise  it  is  impossible,  and  the  equation  cannot 
be  resolved  as  a  quadratic. 

For  example,  reduce  the  following  equation  to  the  quadratic 
form,  if  it  be  possible. 

1.  Given  x4 — 8a#3-f  8aV-f32a3a>- 9a4=0,  to  find  the  values 
of  x  by  quadratics. 

OPERATION. 
a-4—  Sax3  +8aV+32a3*--9a4==0 


This  remainder  can  be  put  into  this  form : 
Now  we  observe  the  original  equation   can  be  written  thus : 


— 9a4=0 
By  putting   #2 — 4ax=y  we  have 

y2 — 8a2y=9a*  a  quadratic. 
Completing  the  square?/2 — 8a2y-j-16a4=25a4 
By  evolution        y — 4a2=±5a2 
Hence  y=9a2     — «2 


QUADRATIC  EQUATIONS.  109 

Or  x2—  4ax=Qa2  or  —a2 

Completing  the  squares2  —  4ax-\-4a2=l3a2  or  3a2 

By  evolution         x  —  2a=±aljl3  or  aj3 
Hence  x  may  have  the  four   following1  values   (2«-{-«>/iy)» 
(2a—aJ~T3),  (2«-f  aj3),  (2a—aj3).     Either  of  which  being 
substituted  in  the  original  equation  will  verify  it. 

2.  Reduce  a73-f-2ax2+5a2^-f  4«3=0  to  a  quadratic. 

As  ihe  highest  power  of  x  is  not  ct'en,  we  must  multiply  by  x 
to  make  it  even.     Then 

x4+2ax3  +5aV-f-4tt3.r  =0 

By  extracting  two  terms  of  the  square  root,  and  observing  the 
remainder,  the  part  that  will  not  come  into  the  root,  we  find  that 


Divide  by  (x2-{-ax)  and  xz-\-ax-}-4a2=Q  a  quadratic. 
3.  Given  x*+2x*—7x2  —  8#-hl2=0,  to  find  the  values  of  x. 
This  equation  may  be  put  in  the  following  form  : 


•Ans.     x=l  or  2,  or  —  2  or  —  3. 

4.  Given  x3  —  8^2-fl9  x  —  12=0,  to  find  the  values  of  x. 

Am.     x=l  or  3  or  4. 

5.  Given  a:4—10^3-|-35a;2—  50#-{-24=0,  to   find   the   values 
of  a?.  JJns.     x=l,  2,  3  or  4. 

6.  Given  x4  —  2x3+a?=132,  to  find  the  values  of  x. 

Jjns.     x=4  or  —  3. 

7.  Given    i/4—  2n/3-f(c2  —  2)t/2-f  2cy=C",   to  find  the    values 

°f  y- 


(Art.  106.)  The  object  of  this  article  is  to  point  out  a  few  lit- 
tle artifices  in  resolving  quadratics,  which  apply  in  particular 
cases  only,  but  which  at  times  may  save  much  labor.  It  is  there- 
fore proper  that  they  should  be  presented,  though  some  minds 
prefer  uniformity  to  facility. 
15 


J70  ELEMENTS  OF  ALGEBRA, 

For  example,  take  equation     (J5)  (Art.  104.) 

1.  y*— -5y=6         Put  2a=5 
Then  y*— 2ay=2a+l. 

Add  a2  to  both  sides  to  complete  the  square,  (Rule  1 .) 

And  i/2— 2ay-f-a2=«2+2a-}-l 
By  evolution,          y — fl=±(a-f-l.) 
Hence,  y=2a-\-l=G  or  — 1 

2.  Given  y* — 7i/=8,  to  find  y.  Jlns.    y=8  or  — 1. 

3.  Given  a:2-j-ll#=26,  to  find  the  values  of  x. 

Assume  2a=ll ;  then  4a-J-4=26. 

Now  put  these  values  in  place  of  the  numerals,  and  complete 
the  square,  and  o;2-{-2rt#-J-a2=a2-|-4a+4. 

By  evolution,  x -\-a=±(a-\-2]  Hence,  x=Z  or — 13. 

4.  Given  x2 — 17^=60,  to  find  the  values  of  x. 

Assume  2a=17;  then  6a-f-9=60 
And  **•— 2«a?+a2=a2+6a-{-9. 
By  evolution,  x — a—±(a-\-3.)     Hence,  rr=20  or  — 3. 

5.  Given  a^-f-19#=92,  to  find  the  values  of  x. 

Assume  2a=19  ;  then  8«+ 16=92 
Putting  these  values  and  completing  the  square  we  have 
a?-}-2ax  +«2=a2+8a4- 1 6 
ar-{-a=±(a-r-4)       or    x=4  or — 23. 

Observe  that  in  the  preceding  equations  we  invariably  put  the 
coefficient  of  the  first  power  of  the  unknown  quantity  equal  2a. 
Then  if  we  find  the  absolute  term  in  the  second  member  of  the 
equation  equal  to  2a-{-  1 

or     4a-f  4 

or     6a-f-  9 

or    8a-}-16 

Or,  in  general,  m2a-\-mz.  That  is,  any  multiplier  of  2«  plus 
the  square  of  the  same  multiplier  equal  to  the  second  member, 
then  the  equation  can  be  resolved  in  this  manner  ;  for  in  fact  one 


QUADRATIC  EQUATIONS.  171 

of  the  roots  of  the  equation  is  this  multiplier  of  2a,  and  the  other 
root  is  ±(2a-j-77i),  m  being  the  multiplier,  and  it  may  represent 
any  number,  integral  or  fractional  ;  but  there  is  no  utility  in  ope- 
rating by  this  method  unless  m  is  an  integer,  and  not  very  large 
To  present  a  case  where  m  is  fractional,  we  give  tie  following 
equation:  xz  —  9#=y,  to  find  the  value  of  x. 

Put    20=9;  then   Ix2a+J=l49»  an(^  ^ie  equati°n   becomes 


Therefore,  x  —  a=±(fl-H).     Hence,  x=  —  £  or  2a+s=9|. 

(Art.  107.)  When  the  roots  of  the  equation  are  irrational  or 
surd,  of  course  this  method  of  operation  will  not  apply  ;  but  we 
can  readily  determine  whether  the  roots  will  be  surd  or  not.  For 
example,  take  the  equation  #2-r"13#=40. 

Put  2a=13;  then  4a-|-4=30     And  60+9=48 
From  this,  we  observe  that  one  of  the  roots  of  the  equation 
lies  between  2  and  3. 

(Art.  108.)  When  the  roots  of  an  equation  are  irrational  or 
surd,  no  artifice  will  avail  us,  and  we  must  conform  to  set  rules  ; 
but  when  the  roots  are  small  integers,  we  can  frequently  find 
some  method  to  avoid  high  numeral  quantities  ;  but  special  artifi- 
ces can  only  be  taught  by  examples,  not  by  precept.  The  follow- 
ing are  given  as  examples  : 

1.  Given  rj-f-9984#=l  60000,  to  find  the  values  of  x. 

Observe  that  9984=10000—16 

Put  2a=10000;  then  320=160000 
These  substitutions  transform  the  equation  to 


•    Completing  the  square  by  (Rule  1)  and 

a*+(2»—  16)a?-r-(a—  8)2=a2-f-16a-f-64 
By  evolution,  x-\-(a  —  8)=db(a-f-8) 
Hence,       #=16  or  —  2a=  —  10000. 

2.  Given  #2+45#=9000,  to  find  the  values  of  x. 

If  we  put  20=45,  the  multiplier  and  its  square,  requisite   to 


172  ELEMENTS  OF  ALGEBRA. 

produce  9000,  is  so  large  that  it  is  not  obvious,  and  of  course 
there  will  be  no  advantage  in  adopting  this  method ;  at  the  same 
time,  we  wish  to  avoid  the  high  numerals  we  must  encounter  by 
any  set  rule  of  solution; 

We  observe  that  45  X  200=9000.     Put  a=45 

Then  x?+ax=2QOa 
Complete  the  square  by  (Rule  2,)  and 
(13?+ «2=a2  +8000 


By  evolution,  2#+a=«ya(a+800)=>/45X845 

Multiply  one  of  the  factors,  under  the  radical,  by  5,  and  divide 
the  other  by  5,  and  the  equivalent  factors  will  be  225  X  169,  both 
squares.  Taking  their  root,  resuming  the  value  of  a,  and  the 
equation  becomes 

2;r+3.15=13.J5 
Drop  3.15  from  both  sides 
And        2a?=10.15     or    #=75,       rfns. 

3.  Given    IQx2  —  225#=225,  to  find  the  values  of  x. 

This  equation  is  found  in  many  of  the  popular  works  on 
algebra,  and  in  several  of  them  the  common  method  of  resolving 
it  may  be  seen. 

Observe  that     225=15X15. 

Put  a=15;  then  a+l  =  16,  and  the  equation  becomes 


Completing  the  square  by  (Rule  2),  and 
4(«+])V—  4(«+l)a2a:  +a4=a4 
By  evolution,     2(a+l)a?  —  «2=<z2+2a 
Transpose  a2  and  divide  by  2,  and  we  have 


Divide  by  (0+1)  and  x=a=\  5,         Jlns 

We  give  one  more  example  of  the  utility  of  representing  nume- 
rals, or  numeral  factors,  by  letters,  in  reducing  the  following 
equation  : 


QUADRATIC  EQUATIONS.  173 

18  ,  81— rr2     r2— 65 
4.  Given   --[— ^— -     — — 

By  examining  the  numerals,  we  find  9,  and  several  multiples 
of  9.  Therefore,  let  a=9,  and  using  a  in  the  place  of  9  the 
equation  becomes 

2a  ,  a2— y?_x*—  65 

"P"1     ^~:  :~~&T~ 

Clearing  of  fractions,  we  have 

16«2-f8a2;r— 8x*=x4— 65^ 

Transposing  all  to  one  side,  and  arranging  the  terms  according 
to  the  powers  of  .T,  we  have 

— 16a«=0 


;—  16a2 
Or      —  «2(x24-8^+16) 

Therefore,  by  (Art.  105,)  the  equation  becomes 


Or        (ar+4)V=«2(a:+4)2 
By  division,     ic2=«2 

And         a;  =±«=±9, 

The  preceding  examples  may  be  of  service  in  reducing  some 
of  the  following 

EXAMPLES. 
1.  Given  a;2-}-  11  a?  =80,  to  find  a?.          ^ns.  a:=5,  car  —  16 


. 

2.  Given   5a?  --  -=2a?-f  -  -  ,   to  find  a?. 

#  —  3  *4 

rfns.     ar=4,  or  —  1 

3.  Given  -  ^+^  ~~"fi"'   to  fill(1  ^  *^n5'     a?=2 


174  ELEMENTS  OF  ALGEBRA. 

72a?  250 

4.  Given   7x-{-- ^-=50,  to  find  x.    Ans.  #=2,  or  - — 

10 — 3x  21 

/6       V     /6       \ 

5.  Given   ( — h*/)-H — hy)=30,  to  find  tne  values  of 

Ans.     y=3  or  2,  or 


_4  £ 

6.  Given  #3-|-7#3=44,  to  find  the  values  of  x. 

Am.     #=±8  or  ±(— 


.  Given  ^-j-ll-j-^/i/2-!-!  1+2=44,  to  find  the  values  of  y. 

fins.     2/=±5  or 
it  -1-7 


it  --  --# 

8.  Given   14-}-2.r  --  =#4-  —  -  —  ,  to  find  the  values  of  x. 

Ans.     #=28  or  9. 

9.  Given   3z?  —  Qx  —  4=80,    to  find  the  values  of  x. 

Am.     x=7  or  —  4. 


10.  Given  =T,  to  find  x.          Ans.     x=4. 


11.  Given  -{-x—  -2=24—  3x,   to  find  the  values 

x  —  3 

of  x.  Ans.     x=G  or  5. 

12.  Given   ---  :  —  =  —  ,   to  find  the  values  of  x. 

x          x*          9 

A)is.     a?—  3  or  y'f. 

-^  _  l  Ox2  4-1 

13.  Given   —  -  —  ^—  r-jr-  =x  —  3,   to  find  the  values  of  x. 

XT—  —  \)X~\~  J 

Ans.     x=l  or  —  28. 

14.  Given  ma?  —  2mxjn=nx*  —  mn,  to  find  x. 

a  Jmn 

Ans.     x=— 


15.  Given   #4H  —  -—=34a?+16,   to   find   the  values  of  x. 
Z 

Exms.  Art.  99.)         Ans   ,r=2,  or  —2,  or  —  8,  or  —  5. 


QUADRATIC  EQUATIONS.  175 

16.  Given   £  ^(J-  V  £3=^,  to  find  the  values  of  *. 

N.B.  Put       -1-=.  fins.     x=S  or  —9. 


17.  Given  y*  —  ty2=y4-8y-f-12,   to  find  the  values  of  y. 
(See  Art.  99.)  Jlns.    y=3  or  —2. 

o 

18.  Given  x  —  1=2+  —  ,   to  find  the  values  of  a:. 

Jx 

Am.  x=4  or  1 


19.  Given   (—-^~  -Y=# — 2,  to  find  the  values  of  x 
\z—Jx*—9/ 


tfns.     x=5  or  3. 


CHAPTER  II. 


Quadratic  Equations,  containing  two  or  more  unknown 
quantities. 

(Art.  108.)  We  have  thus  far,  in  quadratics,  considered  equa- 
tions involving  only  one  unknown  quantity;  but  we  are  now 
fully  prepared  to  carry  our  investigations  farther. 

Two  equations,  essentially  quadratic,  involving  two  unknown 
quantities,  depend  for  their  solution  on  a  resulting  equation  of  the 
fourth  degree. 

This  principle  may  be  shown  in  the  following  manner  : 

Two  equations,  essentially  quadratic,  and  in  the  most  gencrul 
form,  involving  two  unknown  quantities,  may  be  represented 
thus  : 

=0, 


We  do  not  represent  the  first  terms  with  a  coefficient,  as  any 
coefficient  may  be  reduced  to  unity  by  division  ;  and  a,  b,  c,  &c., 
and  a',  b't  &c.  may  represent  the  result  of  such  division  ;  and, 
of  course,  may  be  of  any  value,  whole  or  fractional,  positive  or 

negative. 


176  ELEMENTS  OF  ALGEBRA. 

Arranging  the  terms,  in  the  above  equations,  in  reference  to 
we  have 

a*+(m/+c)x+hf+dy+e    =0,     (1) 

>=Q.     (2) 


13  y  subtracting  (2)  from  (1),  we  have 

'}f+(d—d>)y+e—  e'=0; 


Therefore 


Tliis  expression  for  a?,  substituted  in  either  equation  (1)  or  (2) 
will  give  a  final  equation,  involving  only  one  unknown  quan 
tity,  y. 

But  to  effect  this  substitution  would  lead  to  a  very  complicated 
result ;  and  as  our  object  is  only  to  show  the  degree  to  which 
the  resulting  equation  will  rise,  we  may  observe  that  the  ex- 
pression for  the  value  of  x  is  in  the  form  of  — — — — r— — -.  This 

ry+s 

put  in  either  of  the  equations  (1)  or  (2),  its  square,  or  the  ex- 
pression for  ar2,  will  be  of  the  fourth  degree ;  and  no  term  can 
contain  y  of  a  higher  degree  than  the  fourth. 

Therefore,  in  general,  the  resolution  of  two  equations  of  the 
second  degree,  involving  two  unknown  quantities  depends  upon 
that  of  an  equation  of  the  fourth  degree  involving  one  unknown 
quantity. 

(Art.  109.)  Two  or  more  equations,  involving  two  or  more 
unknown  quantities,  can  be  resolved  by  quadratics,  when  they 
fall  under  one  of  the  following  cases  : 

1st.  One  of  the  equations  only  may  be  quadratic;  the  other 
must  be  simple,  or  capable  of  being  reduced  to  a  simple  form. 

2d.  The  equations  must  be  similar  in  form,  or  the  unknown 
quantities  similarly  involved  or  combined  in  a  similar  manner, 
as  they  combine  in  regular  powers  ;  or, 

3d.  The  equations  must  be  homogeneous  ;  that  is,  the  expo- 
nents of  the  unknown  quantities  must  make  the  same  sum  in 
every  term. 

In  the  first  and  second  oases,  we  eliminate  one  quantity  in 


QUADRATIC  EQUATIONS.  J77 

one  equation  and  substitute  its  value  in  the  other,  or  perform  an 
equivalent  operation,  by  rules  already  explained. 

In  the  third  case,  we  throw  in  a  factor  to  one  unknown  quan- 
tity, to  make  it  equal  to  the  other,  or  assume  it  to  be  so  ;  but 
these  principles  can  only  be  explained  by 

EXAMPLES. 

ax*  -\-bxy  -\-  cyz  =e, 


These  are  homogeneous  equations,  for  the  exponents  of  the 
unknown  quantities  make  the  same  sum  2,  in  every  term.  In 
such  cases,  assume  x=vyi  then  the  equations  become 


__ 


or      = 


a'v*+b'v+c' 

An  equation  involving  the  1st   and  2d   powers   of  v,  and    of 
course,  a  quadratic. 

The  solution  of  this   equation  will   give  v.     Having  v,  we 
have  y2  and  y,  and  from  vy  we  obtain  x. 

For  a  particular  example,  we  give  the  following  : 

1.  Given  £4^--2;™/=12  I  to  find  the  values  of  x  and  y. 

Put   x=vy  ;  then  the  equations  become 
4v2y2—2vy2=12,     or    y*=- 


u2  —  v)2 
And 


f*  Q 

Whence,  —  =  -  =-———.     Dividing  by  2,  then  clearing  of 
Zv  —  v    2-\-3v 

fractions,  we  have 

v=8t>2—  4.v     or     8u2 
12 


178  ELEMENTS  OF  ALGEBRA. 

This  last  equation  gives   v=2   or  —  |. 

8          8 
Omitting  the  negative  value  y*=—  -r-  —  =-=l. 

--  o 


Therefore,  7/=±l,  and  x=vy=±2. 

2.  Given   2x  —  3y=l,   and   2x*-}-xy—  5yz=2Q,   to  find  the 
values  of  x  and  y. 

These  equations  correspond  to  the  first  observation,  one  of 
them  only  being  quadratic,  the  other  simple  ;  and  the  solution 
is  effected  by  finding  the  value  of  x  in  the  first  equation.  Sub- 

stituting that  value   -    —    in   the   2d,  and  reducing,  we  have 
2?/2-{-77/=39,    which  gives  2/=3.     Hence,   x=5. 

3.  Given   &2-\-yz—  x—  #=78,   and   xy+x-\-y=39,   to   find 
the  values  of  x  and  y. 

In  these  equations  x  and  y  are  similarly  involved,  not  equally 
involved  ;  nor  are  the  equations  homogeneous.  In  cases  of  this 
kind,  as  we  have  before  remarked,  a  solution  by  a  quadratic  can 
be  effected,  but  no  general  or  definite  rule  of  operation  can  be 
laid  down  ;  the  hitherto  acquired  skill  of  the  learner,  and  his 
power  of  comparison  to  discern  the  similarity,  will  do  more  than 
any  formal  rules. 

To  resolve  this  example,  we  multiply  the  2d  equation  by  2, 
and  add  the  product  to  the  first  ;  we  then  have 
l5G,     or 


Put  x-}-y=s  ;'  then  s2-f-s=156,  a  quadratic,  which  gives  .9, 
or  x-{-y=l2.  This  value  of  or-J-y,  taken  in  the  second  equa- 
tion, gives  xy=27.  From  this  sum  and  product  of  x  and  y, 
we  find  a?=9  or  3,  and  y=3  or  9. 

Again,  after  we  multiply  the  second  equation  by  2,  if  we  sub- 
tract it  from  the  first,  we  shall  have 


or    (x—y)2—  «(a?+y)=0 
or     (#—  y)2=3Xl2=36 
or    x  —  y—±Q 
But  x+y=l$ 

Hence,         2#=18  or  6,  and  #=9  or  3,  as  before. 


QUADRATIC  EQUATIONS.  179 

(Art.  110.)  There  are  some  equations  to  which  the  foregoing 
observations  do  not  immediately  apply,  or  not  until  after  reduc- 
tions and  changes  take  place.  The  following  is  one  of  them. 


4.  Given  y      >  to  find  the  values  of  x  and  y. 


Here  neither  of  the  equations  is  simple,  nor  are  both  letters 
similarly  involved,  nor  are  the  equations  homogeneous  ;  yet  we 
can  find  a  solution  by  a  quadratic,  because  the  two  equations 
have  a  common  compound  factor,  which  taken  away,  will  bring 
the  equations  far  within  the  limits  or  condition  laid  down  ;  and 
this  remark  will  apply  to  all  problems  that  can  be  resolved  by 
quadratics  not  seemingly  within  the  limits  of  the  three  con- 
ditions. 

12 

From  the  first  of  these  equations,  we  have  y— 


From  the  second,  -  ...    y=         . 

12  18 

Hence,   —~- — =  Divide  the  denominators  by  (#+1) 

X"-\~X      X  + 1 

find  the  numerators  by  6,  and  we  have 

2  3 

~==Z5 — ITTT  a  Qua^ratic  equation. 


Clearing  of  fractions,  and  2#2  —  2x-{-2=3x 

or    2.x2  —  5#=—  -2. 
(Rule  2.)  I6x2—  .#+25=25—16=9. 

We  write  Jl  to  represent  the  second  term.  It  is  immaterial 
what  its  numeral  value  may  be,  as  it  always  disappears  in  taking 
the  root. 

By  evolution,       4#  —  5=  ±3 
Hence,  o?=2   or   5. 

12         12  12       12X4 

*    °r 


The  following  is  of  a  similar  character  : 


180  ELEMENTS  OF  ALGEBRA. 


5    Given  $  **""  2/2~  (*+#)  =  8  ?    to  find    the  values   of  3, 

W$        andy. 


0 

Divide  the  first  equation  by  (x-\-y)  and  x  —  y  —  1—  --  (.#). 

x-{-y 

oo 

Divide   the   second  by   (x-\-y)      and      (x  —  2/)2=--  —  (B). 
Put   x-\-y=s,   and  transpose  minus  1,  in  equation  (.#),  and 
=*  +  l.     By  squaring,  (^-2/)2=^4-y-M. 

00 

Equation  (5)  gives  (x  —  y)2=— 

5 

Therefore,         °V-6+l=??. 

S'2          5  5 

Clearing  of  fractions,  and  transposing  32s,  we  have 

64—  -16s-|-s2=0 

By  evolution,   8  —  s=0    or   s=8.     That  is,  a:+y=8,  which 
value,  put  in  equation  (^j,  gives   x  —  y=%. 
Whence,     x=5     and    2/=3. 

MISCELLANEOUS    EXAMPLES. 

1.  Given   x=2y2   and    5(0?  —  2/)=5,   to  find  the  values  of  x 
and  y.  Jlns.     x=l8  or       12^. 

T/=  3  or  —  2j|. 

2.  Given  2^4-y=22,  and   a?i/+2i/2=120,  to  find  the  values 
of  x  and  ?/.  Jlns.     x=8,  y=G. 

3.  Given    x+y  :  x—  y  :  :  13  :  5,    and    #-|-7/2=25,    to  find 
the  values  of  x  and  y.  Jlns.     x=9,   and  2/=4. 


4.  Given  J^+S/3^8^  ?  to  find  the  values  of  #  and  y. 
+y  =12       5        «^«5.  a?=8  or  4,     =4  or  8 


5.  Given   z*-{-2xy-{-y2=l2Q — 2x — 2y,   and  xy — «/2=8,  to 
find  the  values  of  x  and  y. 

x=Q  or  9,  or  — i 
v=4  or  1,  or  — 3 


6.  Given 


n 


QUADRATIC  EQUATIONS. 
=56 


=56  ^ 

=60  S  to 


va*ues 


or  ±10. 


to  find  the  values   of  x 
and      . 


8.    Given 


C 

s  4  2  i  3?/ 


/  =  68  ? 
?/==i60  S   to 


e  va^ues 
Jlns.     ic= 


In  the  first  four  examples,  one  of  the  equations  is  simple  ;  in 
the  5th  and  6th,  x  and  y  are  similarly  involved  ;  and  the  Oth, 
7th  and  8th  are  homogeneous. 

(Art.  111.)  When  we  have  fractional  exponents,  we  can  re- 
move them,  as  explained  in  (Art.  92.)  ;  but  in  some  cases  it  may 
not  be  important  to  do  so. 

EXAMPLES. 

1.  Given  x*-}-y3'=3x  and  x*-\-y*=x,  to  find  the  values 
of  x  and  y. 

Put  a^=P;   then    x=P*  and   x^=P* 

And^=#;    then   y=Q?   and  y%=Q* 

Now  the  primitive  equations  become 

P3+Q*=3P*,   and   P+Q=P* 
From  the  1st,         #2=(3—  P)P* 
From  the  2d,         Q  =(P—1)P 
By  squaring,          Q2=(P—  1)2P2 

Put  the  two  values  of  Q2  equal  to  each  other,  rejecting  or 
dividing  by  the  common  factor  P2,  and  we  have 


182  ELEMENTS  OF  ALGEBRA. 

or      P2— 2/M-l=3—P 
or      P*—P=2 
Hence,         P  =2  or  1,  and    #=4   or    1,  and  y=8. 

2.  Given   #^+^+2#^+2y^=23,   and   #V3"=36,  to  find 
the  values  of  x  and  y.  Jlns.     #=27  or  8,  y=8  or  27. 

3.  Given   #"-i-i/5=8,  and  #3-f-7/^+#3i/3"=259,  to  find  the 
values  of  #  and  y. 

;  #=125  or    27, 
\  y=  27  or  125. 


4.  Given  x-}-y-\-x+y=2Q1  and   #=8,  to  find  the 
values  of  x  and  y.  Jlns.     #=8,  i/=32. 

#     4#5   33 

5.  Given   —  h—  f==-j-   an(l  #  —  ^2/=5»  to  ^nt^  ^ie  values  of  X 

and  y.  *fl.ns.  a?=9,  y=4 


to 


6.  Given  \  to  ^n(      e  values  of  ^  an(1  2/ 

*=2|,  y=16 


7.  Given   #(ya  +  lJ-^#a  +  i;4-2;r2?/2=55,   and    #ya  + 

t^vt>a=30,   to  find  the  values  of  #  and  y.  (  #=4  or  9 

<  3/=9  or  4 

?.    3.  ±  l 

8.  Given   x3yz=2yz   and   8#3 — 1/^  =  14,  to  find  the  value* 
of  #  and  y.  ^m     C  #=2744  or  8 

^y=9604  or  4 

(Art.  1 12.)  No  additional  principles,  to  those  already  given, 
are  requisite  for  the  solution  of  problems  containing  three  or 
more  unknown  quantities  in  quadratics.  As  in  simple  equations, 
we  must  have  as  many  independent  equations  as  unknown 
quantities. 

As  auxiliary  to  the  solution  of  certain  problems,  particularly  in 
geometrical  progression,  we  give  the  following  problem  : 


QUADRATIC  EQUATIONS.  183 

Given   x+y=s,   and  xy=p,  to  find  the  values  of  x2-j-y2, 
x'-J-y5?  x4+?/4,  and  x5-J-y5,  expressed  in  terms  of  s  and  p. 

Squaring  the  first,      x*-{-2xy-{-y2=s* 
Subtract  twice  the  second,    2xy        —2p 

1st  result,  x*+y2      =s*  —  2p  (A) 

Again, 


=         Ps 


Subtract 


2d  result, 
Square  (A],  and 
Subtract 


x*+y*=s*—  Bps 
x4-\-2x2yz+y4=s*—  4s*p+4pz 
2x*y2        =  2pz 


3d  result, 
Multiply  (A)  by  (#)  and 


x4+y*=s4—4szp-}-2pz 


Subtract 


x2y2(x-\-y)       = 


spz 


4th  result, 


(C) 


x5+y5       =s5  —  5s3p-}-5sp2          (D) 


CHAPTER  III. 
Questions  producing  Quadratic  Equations. 

(Art.  113.)  The  method  of  proceeding  to  reduce  the  question 
into  equations,  is  the  same  as  in  simple  equations  ;  and,  in  fact, 
many  problems  which  result  in  a  quadratic  may  be  brought  out 
by  simple  equations,  by  foresight  and  skill  in  notation.  Others 
again  are  so  essentially  quadratic,  that  no  expedient  can  change 
their  form. 

EXAMPLES. 

1.  A  person  bought  a  number  of  sheep  for  $240.  If  there 
had  been  8  more,  they  would  have  cost  him  $1  a-piece  less. 
What  was  the  cost  of  a  sheep,  and  how  many  did  he  purchase  ? 

240 

Let  x=  the  number  of  sheep  j   then   —  =cost  of  one. 


184  ELEMENTS  OF  ALGEBRA. 

If  he  had   x-\-S   sheep,  _=:cost  of  one. 

240     240    , 

By  the  question,    -  —  —  :  --  1-1 
x      3?-p8 

Clearing  of  fractions,   240or-)-1920=240;r-{-a:2-f-8;c 
Or 


Resolving  gives  #=40  or  —  48  ;  but  a  minus  number  will  not 
apply  to  sheep  ;  the  other  value  only  will  apply  to  the  problem 
as  enunciated. 

This  question  can  be  brought  into  a  simple  equation  thus  :  Let 
x  —  4=  the  number  of  sheep,  then  8  more  would  be  expressed 
by  #4-4,  and  the  equation  would  be 

,™  '  +1.         Put  «=240. 

aH-4 

Then  "    =    «    +1 

x  —  4     x  -|-4 

Clearing  of  fractions,   ax-}-4a—ax  —  4a-\-x*  —  16. 

Transposing,          #2=8a-l-16=8(a+2)=16X  121 
Extracting  square  root,  #=4  X  1  1=44.     Hence,  a:  —  4=40,  the 
number  of  sheep.     Divide  240  by  40,  and  we  have  $6  for  the 
price  of  one  sheep. 

(Art.  114.)  In  resolving  problems,  if  the  second  member  is 
negative  after  completing  the  square,  it  indicates  some  impossi- 
bility in  the  conditions  from  which  the  equation  is  derived,  or  an 
error  in  forming  the  equation,  and  in  such  cases  the  values  of  the 
unknown  quantity  are  both  imaginary. 

2.  For  example,  let  it  be  required  to  divide  20  into  two  such 
parts  that  their  product  shall  be  140. 

Let  r—  one  part,  then  20  —  x=  the  other. 
By  conditions,     20#  —  ^=140 

Or,      x*—  20#=--140 

Completing  the  square,  x2  —  20#-}-100=  —  40 

By  evolution,  x  —  10=^=2^  —  10 

Or,  x  =10 


QUADRATIC  EQUATIONS.  185 

This  result  shows  an  impossibility ;  there  are  no  such  parts 
of  20  as  here  expressed.  It  is  impossible  to  divide  20  into  two 
such  parts  that  their  product  shall  be  over  100,  the  product  of 
10  by  10,  and  so  on  with  any  other  number.  The  product  of 
two  parts  is  the  greatest  possible,  when  the  parts  are  equal. 

3.  Find  two  numbers,  such  that  the  sum  of  their  squares  being 
subtracted  from  three  times  their  product,  11  will  remain;  and 
the  difference  of  their  squares  being  subtracted  from  twice  their 
product,  the  remainder  will  be  14. 

Let  x=  the  greater  number,  and  i/=the  less. 
By  the  conditions,    '3xy — x2 — t/2=ll 
And  2xy — #2-f-i/2=14 

These  are  homogeneous  equations;  therefore, put  x=vy ; 
Then  3w/2— ify2— y2=l  1         (./?) 

And  2vy*—vy+y*=l4         (B) 

Conceive  (./?)  divided  by  (B)  and  the  fraction  reduced,  we  have 
3f— v*— 1^11 
l      14 


Clearing  of  fractions  and  reducing,  we  find 
3u2 — 20u  = — 25. 

5 

A  solution  gives  one  value  of  v,  - 

3 

Put  this  value  in  equation  (.#),  and  we  have 

5y2 v2= !  1 

9       f 

Multiply  by  9.  and  457/2— 25^2— 9y2=llX9, 
Or,     lli/2=llX9, 

yz=  9  or  y=3.    Hence,  x—5. 

A.  A  company  dining  at  a  house  of  entertainment,  had  to  pay 
$3.50 ;  but  before  the  bill  was  presented  two  of  them  went  away ; 
in  consequence  of  which,  those  who  remained  had  to  pay  each 
20  cents  more  than  if  all  had  been  present.  How  many  persons 
dined  ?  Jlns.  7. 

16 


186  ELEMENTS  OF  ALGEBRA. 

5.  There  is  a  certain  number,  which  being  subtracted  from 
22,  and  the  remainder  multiplied  by  the  number,  the  product  will 
be  117.     What  is  the  number  ?  Jlns.     13  or  9. 

6.  In  a  certain  number  of  hours  a  man  traveled  36  miles,  but 
if  he  had  traveled  one  mile  more  per  hour,  he  would  have  taken 
3  hours  less  than  he  did  to  perform  his  journey.     How  many 
miles  did  he  travel  per  hour?  Ans,     3  miles. 

7.  A  person  dies,  leaving  children  and  a  fortune  of  $46,800, 
which,  by  the  will,  is  to  be  divided  equally  among  them  ;  but  it 
happens  that  immediately  after  the  death  of  the  father,  two  of  the 
children  also  die ;  and  if,  in  consequence  of  this,  each  remaining 
child  receive  $1950  more   than  he  or  she  was  entitled  to  by  the 
will,  how  many  children  were  there?  Jlns.     8  children. 

8.  A  gentleman  bought  a  number  of  pieces  of  cloth  for  675 
dollars,  which  he  sold  again  at  48  dollars  by  the  piece,  and  gain- 
ed by  the  bargain  as  much  as  one  piece  cost  him.     What  was 
the  number  of  pieces  I  Jlns.     15. 

This  problem  produces  one  of  the  equations  in  (Art.  10.) 

9.  A  merchant  sends  for  a  piece  of  goods  and  pays  a  certain 
sum  for  it,  besides  4  per  cent,  for  carriage  ;  he  sells  it  for  $390, 
and  thus  gains  as  much  per  cent,  on  the  cost  and  carnage  as  the 
12th  part  of  the  purchase  money  amounted  to.     For  how  much 
did  he  buy  it?  Ans.     $300. 

10.  Divide  the  number  60  into  two  such  parts  that  their  pro- 
duce shall  be  704.  Jins.     44  and  16. 

11.  A  merchant  sold  a  piece  of  cloth  for  $39,  and  gained 
as  much  per  cent,  as  it  cost  him.     What  did  he  pay  for  it? 

Jlns.     $30. 

12.  Jl  and  B  distributed  1200  dollars  each,  among  a  certain 
number  of  persons.     Jl  relieved  40  persons  more  than  B,  and 
B  gave  to  each  individual  5  dollars  more  than  Jl.      How  many 
were  relieved  by  A  and  B?  Jlns.  120  by  Jl,  and  80  by  B. 

This  problem  can  be  brought  into  a  pure  equation,  in  like  man- 
ner as  (Problem  1.) 


QUADRATIC  EQUATIONS.  187 

13.  A  vintner  sold  7  dozen  of  sherry  and  12  dozen  of  claret 
for   £50,  and  finds  that  he  has  sold  3  dozen  more  of  sherry  for 
£10  than  he  has  of  claret  for  £6.     Required  the  price  of  each? 

Ans.     Sherry,  £2  per  dozen;  claret,  £3. 

14.  Ji  set  out  from  C  towards  J9,   and  traveled  7  miles  a 
day.     After  he  had  gone  32  miles,  B  set  out  from  D  towards 
C,  and  went  every  day  ^  of  the   whole  journey  ;  and   after 
he  had  traveled  as  many  days  as  he  went  miles  in  a  day,  he  met 
Jl.     Required  the  distance  from  C  to  2). 

J$ns.     76  or  152  miles  ;  both  numbers  will  answer  the  con- 
dition. 

15.  A  farmer  received  $24  for  a  certain  quantity  of  wheat, 
and  an  equal  sum  at  a  price  25   cents  less  by  the  bushel  for  a 
quantity  of  barley,  which  exceeded  the  quantity  of  wheat  by  16 
bushels.     How  many  bushels  were  there  of  each  ? 

Jlns.     32  bushels  of  wheat,  and  48  of  barley. 

16.  A  and  B  hired  a  pasture,  into  which  Jl  put  4  horses,  and 
B  as  many  as  cost  him  18  shillings  a  week  ;  afterwards  B  put 
in  two  additional  horses,  and  found  that  he  must  pay  20  shillings 
a  week.     At  what  rate  was  the  pasture  hired? 

*flns.     B  had  six  horses  in  the  pasture  at  first,  and  the  price 
of  the  whole  pasture  was  30  shillings  per  week. 

I1?.  Find  those  two  numbers  whose  sum,  product,  and  dif- 
ference of  their  squares,  are  all  equal  to  each  other. 
Ans.  i3±6,  and 


18.  If  a  certain  number  be    divided  by   the   product  of  its 
two  digits,  the  quotient  will  be  2,  and  if  27  be  added  to  the  num- 
ber, the  digits  will  be  inverted.     What  is  the  number? 

JJns.     36. 

19.  It  is  required  to  find  three  numbers,  whose   sum   is  33, 
such  that  the  difference  of  the  first  and  second  shall  exceed  the 
difference  of  the  second  and  third  by  6,  and  the  sum  of  whose 
squares  is  441.  fins.     4,  13,  and  16. 


188  ELEMENTS  OF  ALGEBRA. 

2O.  Find  those  two  numeral  quantities  whose  sum,  product, 
and  sum  of  their  squares,  are  all  equal  to  each  other. 

Jlns.  No  such  numeral  quantities  exist.  In  a  strictly  algebraic 
sense,  the  quantities  are 


21.  What  two  numbers  are  those  whose  product  is  24,  and 
whose  sum  added  to  the  sum  of  their  squares  is  62? 

•fins.     4  and  G. 

22.  It  is  required  to  find  two  numbers,  such  that  if  their  pro- 
duct be  added  to  their  sum  it  shall  make  47,  and  if  their  sum  be 
taken  from  the  sum  of  their  squares,  the  remainder  shall  be  62? 

*ftns.     7  and  5. 

23.  The  sum  of  two  numbers  is  27,  and  the  sum  of  their 
cubes  5103.     What  are  their  numbers?  J%ns.     12  and  15 

24.  The  sum  of  two  numbers  is  9,  and  the  sum  of  their  fourth 
powers  2417.     What  are  the  numbers?  Jlns.     7  and  2. 

25.  The   product  of  two  numbers  multiplied  by  the   sum  of 
their  squares,  is  1248,  and  the  difference  of  their  squares  is  20 
What  are  the  numbers?  Jlns.     6  and  4. 

Let  x-}-y=ihe  greater,  and  x  —  y=the  less. 

26.  Two  men  are   employed  to  do  a  piece  of  work,  which 
they  can  finish  in  12  days.      In  how  many  days  could  each  do 
the  \vork  alone,  provided  it  would  take  one  10  days  longer  than 
the  other?  Ans.     20  and  30  days. 

27.  The  joint  stock  of  two  partners,  Jl  and  B,  was  $1000. 
.tf's  money  was  in  trade  9  months,  and  J5's  6  months  ;  when 
they  shared  stock  and  gain,  Ji  received  $1,140   and  B  $640. 
What  was  each  man's  stock? 

Jlns.     JFs  stock  was  $600  ;  B's  $400. 

28.  A  speculator  from  market,  going  out  to  buy  cattle,  met 
with  four  droves.      In  the  second  were  4  more  than  4  times  the 
square  root  of  one  half  the  number  in  the  first.     The  third  con 
tained  three  times  as  many  as  the  first  and  second.     The  fourth 
was  one  half  the  number  in  the  third  and  10  more,  and  the  whole 


ARITHMETICAL  PROGRESSION.  199 

number  in  the  four  droves  was  1121.     How  many  weie  in  each 
drove?  Ans.     1st,  162  ;  2d,  40 ;  3d,  606 ;  4th,  313 

29.  Divide  the  number  20  into  two  such  parts,  that  the  pro- 
duct of  their  squares  shall  be  9216.  Jlns.     12  and  8. 

30.  Divide  the  number  a  into  two  such  parts  that  the  product 
of  their  squares  shall  be  b. 

rfns.  Greater  part     -f- 


Less  part      |- \(a*-4  jlty*. 


31.  Find  two  numbers,  such  that  their  product  shall  be  equal 
to  the  difference  of  their  squares,  and  the  sum  of  their  squares 
shall  be  equal  to  the  difference  of  their  cubes. 

dns.     ±^5"  and  <l(5±,/5) 


SECTION    V. 

ARITHMETICAL  PROGRESSION. 

CHAPTER  I. 

A  sertes  of  numbers  or  quantities,  increasing  or  decreasing  by 
the  same  difference,  from  term  to  term,  is  called  arithmetical  pro- 
gression. 

Thus,  2,  4,  6,  8,  10,  12,  &c.,  is  an  increasing  or  ascending 
arithmetical  series,  having  a  common  difference  of  2  ;  and  20, 
17,  14,  11,  8,  <fec.,  is  a  decreasing  series,  having  a  common  dif- 
ference of  3. 

(Art.  1 15.)  We  can  more  readily  investigate  the  properties  of 
an  arithmetical  series  from  literal  than  from  numeral  terms.  Thus 
let  a  represent  the  first  term  of  a  series,  and  d  the  common  dif- 
ference. 


190  ELEMENTS  OF  ALGEBRA. 

a,(a-{-d),(aJr2d),(a-\-3d)9(a-\-'ld),  &c.,  represent  an  ascend 
ing  series ;  and 

a,  (a — t?),(a — 2d),(a — 3d), (a — 4cQ,  &c.,  represent  a  descend- 
ing series. 

Observe  that  the  coefficient  of  d,  in  any  term  is  equal  to  the 
number  of  the  preceding  terms. 

The  first  term  exists  without  the  common  difference.  All 
other  terms  consist  of  the  first  term  and  the  common  difference 
multiplied  by  one  less  than  the  number  of  terms  from  the  first. 

Wherever  the  series  is  supposed  to  terminate,  is  the  last  term, 
and  if  such  term  be  designated  by  Z,  and  the  number  of  terms 
by  n,  the  last  term  must  be  a-\-(n — l)d,  or  a — (n — l)d,  accord- 
ing as  the  series  may  be  ascending  or  descending,  which  we  draw 
from  inspection. 

Hence,  L=a-±(n—l)d         (A) 

(Art.  116.)  It  is  manifest  that  the  sum  of  the  terms  will  be  the 
same,  in  whatever  order  they  are  written. 

Take,  for  instance,  the  series  3,    5,    7,   9,  11, 

And  the  same  inverted,  11,    9,    7,    5,    3. 

The  sums  of  the  terms  will  be      14,  14,  14,  14, 14. 

Take  the  series     #,  a-f-  rf,    a-{-2d,    a-}-3d,     a-\-4dt 

Inverted,  «-j-4c/,    a-{-3d,    a-\-2d,   a-}-  d,     a 

Sums  will  be        2a+4rf,  2a+4d,  2a+4d,  2a-\-4d,  2a+4d. 

Here  we  discover  the  important  property,  that,  in  an  arithmeti- 
cal progression,  the  sum  of  the  extremes  is  equal  to  the  sum  of 
any  other  two  terms  equally  distant  from  the  extremes.  Also, 
that  twice  the  sum  of  any  series  is  equal  to  the  extremes,  or 
first  and  last  term  repeated  as  many  times  as  the  series  contains 
terms. 

Hence,  if  S  represents  the  sum  of  a  series,  and  n  the  num- 
ber of  terms,  a  the  first  term,  and  L  the  last  term,  we  shall 
have  2S=n(a+L) 

Or  S=?(a+L)         (B) 

The  two  equations  (.#)  and  (B]  contain  five  quantities,  a,  d, 


ARITHMETICAL  PROGRESSION.  191 

Z,  n,  and  S;  any  three  of  them  being  given,  the  other  two  can 
be  determined. 

Two  independent  equations  are  sufficient  to  determine  two  un- 
known quantities,  (Art.  45,)  and  it  is  immaterial  which  two  are 
unknown  if  the  other  three  are  given. 

By  examining  the  two  equations 

Z=a4-(rc— l)rf        (JS) 

S=%(a+L)  (B) 

We  perceive  that  the  value  of  any  letter,  L  for  example,  can  be 
drawn  from  equation  (B)  as  well  as  from  (A). 

It  can  also  be  drawn  from  either  of  the  equations  after  n  or  a 
is  eliminated  from  them.  Hence,  the  value  of  L  may  take/ot/r 
different  forms.  The  same  may  be  said  of  the  other  letters, 
and  there  being  five  quantities  or  letters  and  four  different 
forms  to  each,  the  subject  of  arithmetical  progression  may  in- 
clude twenty  different  equations.  But  we  prefer  to  make  no 
display  with  these  equations,  believing  they  would  add  dark- 
ness rather  than  light,  as  they  are  all  essentially  included  in  the 
two  equations,  (A)  and  (B),  and  these  can  be  remembered  literal- 
ly and  philosophically,  and  the  entire  subject  more  surely  under- 
stood. 

These  two  equations  are  sufficient  for  problems  relating  to 
arithmetical  series,  and  we  may  use  them  without  modification 
by  putting  in  the  given  values  just  as  they  stand,  and  afterwards 
reducing  them  as  numeral  equations. 

EXAMPLES. 

1  The  sum  of  an  arithmetical  series  is  1455,  the  first  term 
5,  and  the  number  of  terms  30.  What  is  the  common  difference! 

Ans.     3. 

Here  £=1455,  a=5,  n=W.  L  and  d  are  sought. 
Equation  (B)  1455=(5-{-Z)15.     Reduced  Z=92 
Equation  (.#)       92=5-]-29<?.        Reduced  d=3,  fins. 


192  ELEMENTS  OF  ALGEBRA. 

2.  The  sum  of  an  arithmetical  series  is  567,  the  first  term  7, 
and  the  common  difference  2.     What  is  the  number  of  terms? 

Am.     21 

Here  5=567,  a=7,  d=2.     L  and  n  are  sought. 
Equation  (A)     Z=7-f-2n  —  2=5+2n 

Equation  (B)  567=(7+5-f-2n)~=6tt-|-n2 

Or  n2-j-6/i-f-9=576 

n-}-3=24,  or  n=2l,Ans. 

3.  Find  seven  arithmetical  means  between  1  and  49. 

Observe  that  the  series  must  consist  of  9  terms. 
Hence,  o=l,  Z=49,  n=Q. 

Ans.     7,  13,  19,25,  31,  37,  43. 

4.  The  first  term  of  an  arithmetical  series  is  1,  the  sum  of  the 
terms  280,  the  number  of  terms  32.     What  is  the  common  dif- 
ference, and  the  last  term?  Ans.     d=z,  Z=16j. 

5.  Insert  three  arithmetical  means  between  |  and  5* 

Ans.     The  means  are  ~,  T52,  \\ 

6.  Find  nine  arithmetical  means  between  9  and  109. 

Ans.    d--=lO. 

"7.  What  debt  can  be  discharged  in  a  year  by  paying  1  cent 
the  first  day,  3  cents  the  second,  5  cents  the  third,  and  so  on,  in- 
creasing the  payment  each  day  by  2  cents? 

Ans.     1332  dollars  25  cents. 

8.  A  footman  travels  the  first  day  20  miles,  23  the  second,  20 
the  third,  and  so    on,  increasing  the  distance  each  day  3  miles. 
How  many  days  must  he  travel  at  this  rate  to  go  438  miles? 

Ans.     12. 

9.  What  is  the  sum  of  n  terms  of  the  progression  of  1,2,  3, 


1O.  The  sum  of  the  terms  of  an  arithmetical  series  is  950, 
the  common  difference  is  3,  and  the  number  of  terms  25.  What 
is  the  first  term?  Ans.  2. 


ARITHMETICAL  PROGRESSION.  193 

11.  A  man  bought  a  certain  number  of  acres  of  land,  paying 
for  the  first,  $5  ;  for  the  second,  $|  ;  and  so  on.  When  he  came 
to  settle  -he  had  to  pay  $3775.  How  many  acres  did  he  pur- 
chase, and  what  did  it  average  per  acre  ? 

Ans.     150  acres  at  $25£  per  acre. 

Problems  in  Arithmetical  Progression  to  which  the  preceding 
formulas,  (A)  and  (B),  do  not  immediately  apply. 

(Art.  117.)  When  three  quantities  are  in  arithmetical  progres- 
sion, it  is  evident  that  the  middle  one  must  be  the  exact  mean 
of  the  three,  otherwise  it  would  not  be  arithmetical  progression  ; 
therefore  the  sum  of  the  extremes  must  be  double  of  the  mean. 

Take,  for  example,  any  three  consecutive  terms  of  a  series,  as 


and  we  perceive  by  inspection  that  the  sum  of  the  extremes  is 
double  the  mean. 

When  there  are  four  terms,  the  sum  of  the  extremes  is  equal 
to  the  sum  of  the  means,  by  (Art.  116.) 

To  facilitate  the  solution  of  problems,  when  three  terms  are 
in  question,  let  them  be  represented  by  (x  —  y),  x,  (x-\-y],  y  being 
the  common  difference. 

When  four  numbers  are  in  question,  let  them  be  represented 
by  (x  —  3y),  (x  —  y),  (aj+y),  (x-\-3y)i  2y  being  the  common  dif- 
ference. 

So  in  general  for  any  other  number,  assume  such  terms  that 
the  common  difference  will  disappear  by  addition. 

1.  There  are  five  numbers  in  arithmetical  progression,  the 
sum  of  these  numbers  is  65,  and  the  sum  of  their  squares  is 
1005.  What  are  the  numbers  ? 

Let  x=  the  middle  term,  and  y  the  common  difference.  Then 
X  —  2y,  x—y,  x,  x-\-y,  x-\-2y,  will  represent  the  numbers, 
and  their  sum  will  be  5#=65,  or  #=13.  Also,  the  sum  of 
their  squares  will  be 

5#2-f-10<j/2=1005     or     ic2+2»/2=201. 

But  #2=169;  therefore,  2?/2^32,  i/2=16   or  y=i. 

Hence,  the  numbers  are    13  —  8=5,  9,  13,  17  and  21 
17 


194  ELEMENTS  OF  ALGEBRA. 

2.  There  are  three  numbers  in  arithmetical  progression,  their 
sum  is  18,  and  the  sum  of  their  squares   158.     What  are  those 
numbers?  Jlns.     1,  6  and  11 

3.  It  is  required  to  find  four  numbers  in  arithmetical  progres- 
sion, the  common  difference   of  which  shall  be  4,  and  their  coiv 
tinued  product  176985.  Jlns.     15,  19,  23  and  27 

4.  There  are  four  numbers   in  arithmetical   progression,  the 
sum  of  the  extremes  is  8,  and  the  product  of  the   means   15 
What  are  the  numbers?  Jlna.      1,  3,  5,  7. 

*5.  A  person  travels  from  a  certain  place,  goes  1  mile  the  first 
day,  2  the  second,  3  the  third,  and  so  on ;  and  in  six  days  after, 
another  sets  out  from  the  same  place  to  overtake  him,  and  travels 
uniformly  15  miles  a  day.  How  many  days  must  elapse  after 
the  second  starts  before  they  come  together? 

Jlns.     3  days  and  14  days. 
Reconcile  these  two  answers. 

6.  A  man  borrowed  $60  ;  what  sum  shall  he  pay  daily  to  can- 
cel the  debt,  principal  and  interest,  in  60  days;  interest  at  10  per 
cent,  for  12  months,  of  30  days  each? 

Jlns.     $1  and  f  ^  of  a  cent. 

7.  There  are   four  numbers   in  arithmetical  progression,  the 
sum  of  the  squares  of  the  extremes  is  50,  the  sum  of  the  squares- 
of  means  is  34  ;  what  are  the  numbers?  J$ns.     1,  3,  5,  7 

§.  The  sum  of  four  numbers  in  arithmetical  progression  is 
24,  their  continued  product  is  945.  What  are  the  numbers '' 

Jlns.     3,  5,  7,  9. 

9.  A  certain  number  consists  of  three  digits,  which  are  in 
arithmetical  progression,  and  the  number  divided  by  the  sum  of 
its  digits  is  equal  to  26 ;  but  if  198  be  added  to  the  number  its- 
digits  will  be  inverted  What  is  the  number  1  fins.  234 


GEOMETRICAL  PROGRESSION.  195 

CHAPTER  II. 

GEOMETRICAL  PROGRESSION. 

(Art.  118.)  When  numbers  or  quantities  differ  from  each 
othfr  by  a  constant  multiplier  in  regular  succession,  they  consti- 
tute a  geometrical  series,  and  if  the  multiplier  be  greater  than 
unity,  the  series  is  ascending  ;  if  it  be  less  than  unity,  the  series 
is  descending. 

Thus,  2  :  6  :  18  :  54  :  162  :  486,  is  an  ascending  series,  the 
multiplier,  called  the  ratio,  being  three  ;  and  81  :  27  :  9  :  3  :  1  : 
l  :  i,  &c.,  is  a  descending  series,  the  multiplier  or  ratio  being  |. 

Hence,  a  :  ar  :  ar*  :  «r3  :  ar4  :  ar5  :  w6  :  &c.,  may  represent 
any  geometrical  series,  and  if  r  be  greater  than  1,  the  series  is 
ascending,  if  less  than  1,  it  is  descending. 

(Art.  119.)  Observe  that  the  first  power  of  r  stands  in  the 
2d  term,  the  2d  power  in  the  3d  term,  the  third  power  in  the 
4th  term,  and  thus  universally  the  power  of  the  ratio  in  any 
term  is  one  less  than  the  number  of  the  term. 

The,  first  term  is  a  factor  in  every  term.  Hence  the  10th 
term  of  this  general  series  is  ar9.  The  17th  term  would  be  ar16. 
The  rath  term  would  be  arn~l. 

Therefore,  if  n  represent  the  number  of  terms  in  any  series, 
and  L  the  last  term,  then  L=ai**~l  (1) 

(Art.  120.)  If  we  represent  the  sum  of  any  geometrical  series 
by  5,  we  have 

5=a-j-ar+cfr2-}-«r34-  &c.  .  .  arn-2+arn~l. 
Multiply  this  equation  by  r,  and  we  have 

r5=ar-}-tfr2-f  ar3-f-  &c.  arn~l-\-arn. 
Subtract  the  upper  from  the  lower,  and  observe  thai 
Zr=arn;  then  (r  —  l)s=Lr  —  a. 


Therefore,  8=-         (2) 

As  these  two  equations  are  fundamental,  and  cover  the  whole 
subject  of  geometrical  progression,  let  them  be  brought  together 
for  critical  inspection. 


196  ELEMENTS  OF  ALGEBRA. 


(1),         S=  (2). 

These  two  equations  furnish  the  rules  given  for  the  operations 
in  common  arithmetic. 

Here  we  perceive  five  quantities,  0,  ?*,  n,  L  and  S,  and  any 
three  of  them  being  given  in  any  problem,  the  other  two  can  be 
determined  from  the  equations. 

To  these  equations  we  may  apply  the  same  remarks  as  were 
made  to  the  two  equations  in  arithmetical  progression  (Art.  116.) 

Equation  (2),  put  in  words,  gives  the  following  rule  for  the 
sum  of  a  geometrical  series  ; 

RULE.  Multiply  the  last  term  by  the  ratio,  .and  from  the 
vroduct  subtract  the  first  term,  and  divide  the  remainder  by  the 
ratio  less  one. 

EXAMPLES    FOR    THE   APPLICATION   OF   EQUATIONS 
(1)    AND    (2). 

1.  Required  the  sum  of  9  terms  of  the  series,  1,  2,  4,  8,  16, 
&c.  Ans.  511. 

3.  Required  the  8th  term  of  the  progression,  2,  6,  18,  54, 
&c.  Ans.  4374. 

3.  What  is  the  sum  of  ten  terms  of  the  series  1,  f,  J,  &c.  ? 

•*»•    WAV 

4.  Required  two  geometrical  means  between  24  and  192. 

N.  B.  When  the  two  means  are  found,  the  series  will  consist 
of  four  terms  ;  the  first  term  24  and  the  last  term  192. 

By  equation     (1)     L=arn~\ 
Here  a—  24,  Z=192,  n=4,  and  the  equation  becomes 

192=24r3  or  r=2. 
Hence,  48  and  96  are  the  means  required. 

5.  Required  7  geometrical  means  between  3  and  768. 

tins.     6,  12,  24,  48,  96,  192, 

6.  The  first  term  of  a  geometrical  series  is  5,  the  last  term 
1215,  and  the  number  of  terms  6.     Wliat  is  the  ratio  1     fins.  3. 


GEOMETRICAL  PROGRESSION.  J97 

7".  A  man  purchased  a  house,  giving  $1  for  the  fiist  door,  $2 
for  the  second,  $4  for  the  third,  and  so  on,  there  being  10  doors. 
What  did  the  house  cost  him  1  tins.  $1023. 

(Art.  121.)  By  Equation  (2),  and  the  Rule  subsequently  given, 
we  perceive  that  the  sum  of  a  series  depends  on  the  first  and  last 
terms  and  the  ratio,  and  not  on  the  number  of  terms ;  and 
whether  the  terms  be  many  or  few,  there  is  no  variation  in  the 
rule.  Hence,  we  may  require  the  sum  of  any  descending  series, 
as  1,  a»  i,  \,  &c.,  to  infinity,  provided  we  determine  the  LAST 
term.  Now  we  perceive  the  magnitude  of  the  terms  decrease 
as  the  series  advances ;  the  hundredth  term  would  be  e'xtremely 
small,  the  thousandth  term  very  much  less,  and  the  infinite  term 
nothing  ;  not  too  small  to  be  noted,  as  some  tell  us,  but  absolutely 
nothing. 

Hence,  in  any  decreasing  series,  when  the  number  of  terms 
is  conceived  to  be  infinite,  the  last  term,  Z,  becomes 

0,  and  Equation  (2)  becomes  s=——-- 

By  .change  of  signs  s=- -• 

This  gives  the  following  rule  for  the  sum  of  a  decreasing  infi- 
nite series : 

RULE.  Divide  the  first  term  by  the  difference  between  unity 
and  the  ratio. 

EXAMPLES. 

1.  Find  the  value  of  1,  £,  T9¥,  &c.,  to  infinity. 

o=l,  r=\.  JLns.    4 

2.  Find  the  exact  value  of  the  decimal  .3333,  &c.,  to  infinity. 

rfns.     5. 

This  may  be  expressed  thus  :  T\+Tf  ¥,  <fec.     Hence. 

3.  Find  the  value  of  .323232,  &c.,  to  infinity. 

«=T3«2o»  «r=T<nr2oo ;  therefore  r=T±-0.  Ans.     §| 

4.  Find  the  value  of  .777,  &c.,  to  infinity.  *ftns.     £. 


J98  ELEMENTS  OF  ALGEBRA. 

5.  Find  the  value  of  f  :  1  :  J  :  /3,  &c.   to  infinity. 

tins.     4 ' 


6.  Find  tlie  value  of  5  :  f  :  £ ,  &c.  to  infinity.         Jlns.     7| 

7.  Find  the  value  of  the  series  1,  ^,  &c.,  to  infinity  ?  Am.  |. 

8.  What  is  the  value  of  the  decimal    .71333,  &c.,  to  infinity? 

•On*.    Hi- 
O.  What  is  the  value  of  the  decimal  .212121,  &c.,  to  infinity? 

•*>«.  A- 

(Art.  122.)  If  we  observe  the  general  series,  (Art.  11 8.)  a:  an 
ar2  :  ar3  :  ar4,  &c.,  we  shall  find,  by  taking  three  consecutive 
terms  anywhere  along  in  the  series,  that  the  product  of  the  ex- 
tremes will  equal  the  square  of  the  mean.  Hence,  to  find  a 
geometrical  mean  between  two  numbers,  we  must  multiply  them 
together,  and  take  the  square  root.  If  we  take  four  consecutive 
terms,  theproduct  of  the  extremes  will  be  equal  to  the  product 
of  the  means. 

(Art.  123.)  This  last  property  belongs  equally  to  geometrical 
proportion,  as  well  as  to  a  geometrical  series,  and  the  learner  must 
be  careful  not  to  confound  proportion  with  a  series. 

a:  ar  :  :  b  :  br,  is  a  geometrical  proportion,  not  a  continued 
series.  The  ratio  is  the  same  in  the  two  couplets,  but  the  mag- 
nitudes a  and  b,  to  which  the  ratio  is  applied,  may  be  very  dif- 
ferent. 

We  may  suppose  a  :  ar  two  consecutive  terms  of  one  series, 
and  b  :  br  any  two  consecutive  terms  of  another  series  having 
the  same  ratio  as  the  first  series,  and  being  brought  together  they 
form  a  geometrical  proportion.  Hence,  the  equality  of  the  ra- 
tio constitutes  proportion. 

To  facilitate  the  solution  of  some  difficult  problems  in  geomet- 
rical progression,  it  is  desirable,  if  possible,  to  express  several 
terms  by  two  letters  only,  and  have  them  stand  symmetrically. 

Three  terms  maybe  expressed  by  x  :  ,Jxy  :  y,  or  by  xz :  xy  : 
y\  as  the  product  of  the  extremes  is  here  evidently  equal  to  the 
square  of  the  mean. 

To  express  four  terms  with  x  and  y  symmetrically,  we  at  first 


GEOMETRICAL  PROGRESSION.  199 

write  P  :  x  \  :  y  :  Q.  The  firstthree  being  in  geometrical  progres- 

y&  y* 

sion,  gives  Py=x2orP= — .  In  the  same  manner,  we  find  $=— - 

s?  y2 

And  taking  these  values  of  P  and  Q  we  have— : x:: y:  —  to  ren- 

y        x 

resent  four  numbers  symmetrically  with  two  letters. 

Taking  three  numbers  as  above,  and  placing  them  between  P 
and  Q,  thus,  P  :  xz  :  xy  :  y2 :  £,  we  have  five  numbers  ;  and 

3? 

by  reducing  P  and  Q  into  functions  of  x  and  ?/,  we  have—:  #*  : 

V3 
xy  :  yz  :  y—,  for  five  terms  symmetrically  expressed. 

X 

3?    z2  y2    y* 

Six  numbers  thus,  —„:—    :  x  : :  y  :  —  :  ^ 

t/2    y  xx* 

Sometimes  we  may  more  advantageously  express  unknown 
numbers  in  geometrical  proportion  by  x,  xy,  xy2,  <fec. ;  x  being 
the  first  term,  and  y  the  ratio. 

HARMONICAL    PROPORTION. 

(Art.  124.)  When  three  magnitudes,  a,  ft,  c,  have  the  relation 
of  a :  c  :  :  a — b  :  b — c  ;  that  is,  the  first  is  to  the  third  as  the  dif- 
ference between  the  first  and  second  is  to  the  difference  between 
the  second  and  third,  the  quantities  a,  ft,  c,  are  said  to  be  in  har- 
monical  proportion. 

(Art.  125.)  Four  magnitudes  are  in  harmonical  proportion 
when  the  first  is  to  the  fourth  as  the  difference  between  the  first 
and  second  Is  to  the  difference  between  the  third  and  fourth. 
Thus,  «,  ft,  c,  d,  are  in  harmonical  proportion  when  a  :  d  :  : 
a — ft  :  c — d,  or  when  a  :  d :  :  b — a  :  d — c. 

An  harmonical  mean  between  two  numbers  is  equal  to  twice 
their  product  divided  by  their  sum.  For  a  :  x  :  b  representing 
three  numbers  in  harmonical  proportion,  we  have  by  the  definition, 
(Art.  124.)  a  :  ft  :  :  a — x  :  x—b. 

Therefore,  ax — ab=ab — bx  or  x= — r-y. 


200  ELEMENTS  OF  AUJEHKA. 

1.  Find  the  harmonica!  mean  between  6  and  12.        JJns.     8. 

2.  Find  the  third  harmonical  number  to  234  and  144. 

rfns.     104. 

3.  Find  the  fourth  harmonical  proportion  to  the  numbers  24  : 
16  :  4.  Am.     3, 

4U  There  are  four  numbers  in  harmonical  proportion,  the  firs' 
is  16,  the  third  3,  the  fourth  2.     The  second  is  lost  ;  find  it. 

Ans.     8 

PROBLEMS  IN  GEOMETRICAL   PROGRESSION,  AND 
HARMONICAL   PROPORTION. 

1.  The   sum  of  three  numbers  in  geometrical  progression  is 
26,  and  the  sum  of  their  squares  364.      What  are  the  numbers  ? 
Let  the  numbers  be  represented  by  x  :  ,Jxy  :  y. 
Then   x+Jxy+y  =  26=a  (1) 

And      x2-!-    ^4-2/2=364=6  (2) 

Transpose,  ,Jxy  in  Equation  (I)  and  square,  we  have 

a*+2xy+if=(fi—2ajxy+xy          (3) 
Or  x*-\-  xy-{-y2=a~2alJxy  (4) 

The  left  hand  members  of  Equations  (2)  and  (4)  are   equal  ; 

hence,  — 

a2  —  2ajxy=b. 


Therefore,  7^==6      (5) 


This  gives  the  second  term  of  the  progression,  and  now  from 
equations  (1)  and  (5)  we  find  #=2,  y=I8,  and  the  numbers 
are  2,  6,  18. 

2.  The  sum  of  four  numbers  in  geometrical  progression  is  15, 
(«),  and  the  sum  of  their  squares  85  (6).  What  are  the 
numbers  ? 

Let  the  numbers  be  represented  as  in  (Art.  123.) 


Then   — 


GEOMETRICAL  PROGRESSION.  201 

Assume         x-{-y=s 
xy=p 


Then  by  (Art.  112.) 

X'  "'"* 

And     — • 

y 

And  x*-\-y*=s8 — 3sp 

Transposing  (x-\-y)  in  Equation  (1),  and  (x2-\-y2)  in  Equation 
(2),  we  have 


±=a—s  (3)  and     +=b--#+2p          (4) 
y   *x  y2     xz 

Square  (3)  and  transpose  2xy  or  2p  and 

|'+|4=(«-*)2-2/>  (5) 

The  left  hand  members  of  equations  (4)  and  (5)  are  equal,  there 
fore,  (ar-8)*—2p=b—  s2  +2p 

Or  a2—  2as-{-2sz—  4p=b  (6) 

Clear  equation  (3)  of  fractions,  and  x3Jry3—ap  —  ps. 

s3 

That  is,  6-3  —  3sp—an  —  ps  or  »=  —  :  -         (7} 

a+2s 

Put  this  value  of  p  in  equation  (6)  and  reduce,  we  have, 


Or  as*+bs=:- 

2 

Taking  the  given  values  of  a  and  b  we  have 

15s2-|-856'=70X15 

Or     3*2-{-17s=210,  an  equation  which  gives  s=6 
Put  the  values  of  a  and  s  in  equation  (7),  and  p=S 
That  is,  x-\-y=Q,  and  xy=8,  from  which  we  find  #=2,  am! 
=4  ;  therefore  the  required  numbers  are 

1,2,4  and  8,     Ans 

3.  The  arithmetical  mean  of  two  numbers  exceeds  the  geo 


202  ELEMENTS  OF  ALGEBRA. 

metrical  mean  by  13,  and  the  geometrical  mean  exceeds  the  har- 
monical  mean  by  12.     What  are  the  numbers? 

Let  x  and  y  represent  the  numbers. 
Then   |  (#+?/)=    the  arithmetical  mean,  >Jxi/=    the  geome- 

trical mean,  (Art.  112.)  and    —  r^-=   the  harmonical  mean. 

v+y 

Let  o=12; 

Then,  by  the  question,    i(x-\-y)  =  iJxy  -\-a-\-  1  (1) 

«  (2) 


By  our  customary  substitution,  these  equations  become 
5*=V/>+«-fl  (3) 

And         <//>=T+a  (4) 

o 

Take  the  ralue  of  s  from  equation  (3)  and  put  it  into  equation 
(4),  dividing  me  numerator  and  the  denominator  by  2,  and  we 
have  .  p  . 


Clearing  of  fractions,  we  shall  have 


Drop  equals,  and   Jp=(a-\-l}a      (6) 

Put  this  value  of  Jp  in  equation  (3)  and  we  have 


Or         5=2(0+ 1)2  (7) 

For  the  sake  of  brevity,  put  (a-j-l)=6;  squaring  equation 
(6)  and  restoring  the  values  of  s  and  p  in  equations  (6)  and  (7), 
and  we  have  xy=azb*  (Jl) 

x+y=2b*        (JS) 
Square  (S)  and 


Subtract  4 

/  a\ 
times    ^ 


And    xz—  2^-l-i/8=4&2(62—  «2)=462(6+«)(6—  a)  (C) 


GEOMETRICAL  PROGRESSION.  803 

As  «=12  and  b=l3,  b-\- a=25,  and  b— a=l. 
Therefore,  (C)  becomes  (a?— 2/)2=4&2X25X  1. 

By  evolution,     a? — y=2bX5 

Equation  (£)     #-{-i/=2£2 

By  addition  2# 


Or  x=  62-f-  5&=(6-f  5)6=18X13=234 

By  subtraction,       2y=2b2—  Wb 

y=  b2 —  56=(6— 5)6=  8X13=104. 
A  more  brief  solution  is  the  following : 
Let   x — y   and   x-\-y  represent  the  numbers. 
Then   x=   the  arithmetical  mean,    >Jx2 — yz=   the  geometri- 
cal mean,  (Art.  112),  and —  =    the  harmonical  mean.     By 

the  question, 


x~  I3  =  jx2— y2  (1),  and  - — ^-4-12  =  7^— y2          (2) 
The  right  hand  members  of  equations  (1)  and  (2)  being  the 

y2 yZ 

same,  therefore,    ^--{-I2=x — 13. 

x 

By  reduction,    y2=25x. 
Put  this  value  of  yz  in  equation  (1),  and  by  squaring 

#2— 26z-l-(13)2=z2— 25*,  or  a;=(13)2=169. 
Hence,  i/=65,  and  the  numbers  are  104  and  234. 

4.  Divide  the  number  210  into  three  parts,  so  that  the  last 
shall  exceed  the  first  by  90,  and  the  parts  be  in  geometrical  pro- 
gression, dns.     30,  60,  and  120. 

5.  The  sum  of  four  numbers  in  geometrical  progression  is 
30  ;  and  the  last  term  divided  by  the  sum  of  the  mean  terms  is 
1|.     What  are  the  numbers  ?  Ans.     2,  4,  8,  and  16. 

6.  The  sum  of  the  first  and  third  of  four  numbers  in  geo- 
metrical progression  is  1 48,  and  the  sum  of  the  second  and 
fourth  is  888.  What  are  the  numbers  ? 

Ans.     4,  24, 144,  and  864. 


204  ELEMENTS  OF  ALGEBRA. 

7".  It  is  required  to  find  three  numbers  in  geometrical  progres- 
sion, such  that  their  sum  shall  be  14,  and  the  sum  of  their  squares 
84.  Jlns.  2,  4,  and  8. 

8.  There  are  four  numbers   in   geometrical  progression,   the 
second  of  which  is  less  than  the  fourth  by  24  ;    and  the  sum  of 
the  extremes  is  to  the  sum  of  the  means,  as  7  to  3.      What  are 
the  numbers  ?  Jlns.     1,  3,  9  and  27. 

9.  The  sum  of  four  numbers  in  geometrical  progression  is 
equal  to  the  common  ratio  -}-!»  and  the  first  term  is  y1^.      What 
are  the  numbers  ?  fins.     T1^,  T\,  T\,  T£. 

10.  The  sum  of  three  numbers  in  harmonical  proportion  is 
26,  and  the  product  of  the  first  and  third  is  72.      What  are  the 
numbers  ?  Jlns.     12,  8,  and  6 

11.  The  continued  product  of  three  numbers  in  geometrical 
progression  is  216,  and  the  sum   of  the  squares  of  the  extremes 
is  328.     What  are  the  numbers  ?  tins.     2,  6,  18. 

12.  The  sum  of  three  numbers  in  geometrical  progression  is 
13,  and  the  sum  of  the  extremes  being  multiplied  by  the  mean, 
the  product  is  30.     What  are  the  numbers  ? 

1,  3,  and  9 


13.  There  are  three  numbers  in  harmonical  proportion,  the 
sum  of  the  first  and  third  is  18,  and  the  product  of  the  three  is 
576.     What  are  the  numbers  ?  Jlns.     6,  8,  12. 

14.  There  are  three  numbers  in  geometrical  progression,  the 
difference  of  whose  difference  is  6,  and  their  sum  42,     What 
are  the  numbers?  Jlns.     6,  12,24. 

15.  There  are  three  numbers  in  harmonical  proportion,  the 
difference  of  whose  difference  is  2,  and  three  times  the  product 
of  the  first  and  third  is  216.     What  are  the  numbers  ? 

Jlns.     6,  8,  and  12. 

16.  Divide   120.  dollars    between   four   persons,   in   such   a 
way,  that  their  shares  may  be  in  arithmetical  progression  ;  and 
if  the   second  and   third  each  receive  12  dollars  less,  and  tho 


GEOMETRICAL  PROPORTION.  205 

fourth  24  dollars  more,  the  shares  would  then    be    in   geometri- 
cal progression.     Required  each  share. 

Jlns.  Their  shares  were  3,  21,  39,  and  57,  respectively. 

17.  There  are  three  numbers  in  geometrical  progression, 
whose  sum  is  31,  and  the  sum  of  the  first  and  last  is  26.  What 
are  the  numbers  ?  Jlns.  1,  5,  and  25. 

18  The  sum  of  six  numbers  in  geometrical  progression  is 
189,  and  the  sum  of  the  second  and  fifth  is  54.  What  are  the 
numbers  ?  rfns.  3,  6,  12,  24,  48  and  96. 

19.  The  sum  of  six  numbers  in  geometrical  progression  is 
189,  and  the  sum  of  the  two  means  is  36.     What  are  the  num- 
bers? Am.     3,  6,  12,  24,  48  and  96. 
CHAPTER  III. 
PROPORTION. 

(Art.  176.)  We  have  given  the  definition  of  geometrical  pro- 
portion in  (Art.  41.)  and  demonstrated  the  most  essential  prop- 
erty, the  equality  of  the  products  between  extremes  and  means. 
We  now  propose  to  extend  our  investigations  a  little  farther. 

Proportion  can  only  exist  between  magnitudes  of  the  same 
kind,  and  the  number  of  times  and  parts  of  a  time,  that  one 
measures  another,  is  called  the  ratio.  Ratio  is  always  a  num- 
ber, and  not  a  quantity. 

(Theorem  1.)  If  two  magnitudes  have  the  same  ratio  as 
two  others,  the  first  two  as  numerator  and  denominator  may 
form  one  member  of  an  equation  ;  and  the  other  two  magnitudes 
as  numerator  and  denominator  will  form  the  other  member. 

Let  A  and  B  represent  the  first  two  magnitudes  and  r  their  ratio. 

Also  C  and  D  the  other  two  magnitudes,  and  r  their  ratio. 

Then,^=r  and    °=r    Therefore,  (Ax.  7)  —  =£ 

(Theorem  2.)  Magnitudes  which  are  proportional  to  tlie 
same  proportionals,  are  proportional  to  each  other. 

Suppose  a  :  b  :  :  P  :  Q  \       Then  we  are  to  prove  that 
and      c:  d::  P:  Q  V  a:b::c:d 

and     x  :  y  : :  P  :  Q  )  and  a :  b  : :  x :  y,  &c. 


200  ELEMENTS  OF  ALGEBRA. 

b      Q 
From  the  first   proportion,  -=73 

From  the  second,  — =  ~ 

Hence,  (Ax.  7)  —  =  —  or  a  :  b  : :  c  :  d 

In  the  same  manner  we  prove  a  :  b  :  :  x  :  y 
And         c  :  d  :  :  x  :  y 

(Theorem  3.)  If  four  magnitudes  constitute  a  proportion, 
the  first  will  be  to  the  sum  of  the  first  and  second,  as  the 
third  is  to  the  sum  of  the  third  and  fourth. 

By  hypothesis,  a  :  b  :  :  c  :  d  ;^then  we  are  to  prove  that 
a  :  a-\-b  :  :  c  :  c-\-d. 

By  the  given  proportion,— =-.      Add  unity  to  both  members, 

then  reducing  to  the  form  of  a  fraction  we  have = . 

a  c 

Throwing  this  equation  into  its  equivalent  proportional  form,  we 
have'  a  :  a+b  :  :  c  :  c+d. 

N.  B.    In  place  of  adding  unity,  subtract  it,  and  we  shall  find 

that  a  :  a-b  :  :  c  :  c-d. 

or  a  :  b — a  :  :  c  :  d — c. 

(Theorem  4.)  If  four  magnitudes  be  proportional,  the  sum 
of  the  first  and  second  is  to  their  difference,  as  the  sum  of 
the  third  and  fourth  is  to  their  difference. 

Admitting  that  a  :  b  :  :  c  :  d,  we  are  to  prove  that 
a-\-b  :  a — b  :  :  c-{-d  :  c — d 

X 

From  the  same  hypothesis,  (Theorem  3.)  gives 

a  :  a-\-b  :  :  c  :  c-\-d 
And  a  :  a — b  :  :  c  :  c — d 

Changing  the  means,  (which  will  not  affect  the  product  of-  the 
extremes  and  means,  and  of  course  will  not  destroy  proportion- 
ality,) and  we  have, 


GEOMETRICAL  PROPORTION.  207 

arc::  «-}-&  :  c-\-d 

a  i  c  :  :  a — b  :  c — d 

Now  by  (Theorem  2.)         a-{-b  :  c-\-d  :  :  a — b  :  c — d 
Changing  the  means,  a-\-b  :  a — b  :  :  c-\-d  :  c — d 

(Theorem  5.)    If  four   magnitudes    be   proportional,    like 
powers  or  rootsof  the  same,  will  be  proportional. 

Admitting  a  :  b  :  :  c  :  d,  we  are  to  show  that 


n  n  n  n 

an  :  bn  : :  cn  :  er,  and  a     :   b     :  :    c     :  d 

By  the  hypothesis,   ~j-=-7'     Raising  both  members  of  this 
equation  to  the  nth  power,  and 

an     cn 


Changing  this  to  the  proportion   an   :  bn  : :  cn   :  dn 
Resuming  again  the  equation   T=-y»  and  taking  the  nth  root 


of  each  member,  we  have     — j-  =—7-.     Converting  this  equation 

r~**~        j"»~ 

b        d 
into  its  equivalent  proportion,  we  have 

JL        _L  i          i 

n          ,  n  n  ,~n~ 

a      :  b      : :  c      id. 

Now  by  the  first  part  of  this  theorem,  we  have 

m  jrn  m  m 

an    :  b"    :  :  c n   :  d"  ,    m  representing  any 
power  whatever,  and  n  representing  any  root. 

(Theorem  6.)  If  four  magnitudes  be  proportional,  also 
four  others,  their  compound,  or  product  of  term  by  term,  will 
form  a  proportion. 


208  ELEMENTS  OF  ALGEBRA. 

Admitting  that  a  :    b  :  :  c     .  d 

And  x  :    y  :  :  m  :  n 

We  are  to  show  that,        ax  :  by  :  :  me  :  nd 
From  the  first  proportion,  T=y 

From  the  second,  -= — 

y    n 

Multiply  these  equations,  member  by  member,  and 

ax_mc 
by     nd 
Or        ax  :  by  : :  me  :  nd. 

The  same  would  be  true  in  any  number  of  proportions. 
(Theorem  7.)   Taking  the  same  hypothesis  as  in  (Theorem 
6.)  we  propose  to  show,  that  a  proportion  may  be  formed  by  di- 
viding one  proportion  by  the  other,  term  by  term. 
By  hypothesis,  a  :  b  :  :  c  :  d 
And       x  :  y  :  :  m  :  n 

Multiply  extremes  and  means,    ad=bc        (1) 
And      nx=my       (2) 

Divide  (1)  by  (2),  and     -X  -=-  X  - 
x     n    m    y 

Convert  these  four  terms,  which  make  two  equal  products,  into 
a  proportion,  and  we  shall  have 

a  ^  b         c   f  d 
x  '  y        m  '  n 

By  comparing  this  with  the  given  proportions,  we  find  it  com- 
posed of  the  quotients  of  the  several  terms  of  the  first  propor- 
tion divided  by  the  corresponding  term  of  the  second. 

(Theorem  8.)  If  four  magnitudes  be  proportional,  we  may 
multiply  the  first  couplet  or  the  second  couplet,  the  antecedents 
or  the  consequents,  or  divide  them  by  the  same  factor,  and  the 
results  will  be  proportional  in  every  case. 


GEOMETRICAL  PROPORTION.  209 

Suppose a  :  b  : :  c  :  d 

Multiply  ex.  and  means,  and  •  •  •  •  ad=bc  (I) 

Multiply  this  eq.  by  ?n,  and  ....         mad—mbc 

Now,  in  this  last  equation,  ma  may  be  considered  as  a  single 
term  or  factor,  or  md  may  be  so  considered.  So,  in  the  second 
member,  we  may  take  mb  as  one  factor,  or  me.  Hence  we  may 
convert  this  equation  into  a  proportion  in  four  different  ways. 

Thus,  as .  . ma  :  mb  : :  c     :  d 

or    as a     :  b     ::  me  :  md 

or   as ma  :  b     : :  me  :  d 

or    as a     :  mb  ::  c     :  md 

If  we  resume  the  original  equation  (1),  and  divide  it  by  any 
number,  m,  in  place  of  multiplying  it,  we  can  have,  by  the  same 
course  of  reasoning, 


a 

b 

:  :     c 

:     d 

m 

m 

a 

:     b 

c 

d 

m 

m 

a 

:     b 

c 

:      d 

m 

m 

b 

d 

a 

•    

:  :     c 

•    __ 

m 

771 

The  following  examples  are  intended  to  illustrate  the  practi- 
cal utility  of  the  foregoing  theorems  : 


EXAMPLES. 

!•  Find  two  numbers,  the  greater  of  which  shall  be  to  the 
less,  as  their  sum  to  42  ;  and  as  their  difference  is  to  6. 

Let  a?=the  greater,    y=the  less. 
18 


210  ELEMENTS  OF  ALGEBRA. 

(  x  :  y  :  :  x-\-y  :42         (1) 
fhen.pcr  quest™,  j  x  .  *  .  .  ^  :    6         fo 


(Theorem  2.)  x-\-y  :  42  :  :  #—  y  :    6 

Changing  the  means     x-}-y  '  x  —  y  :  :  42  :    6 
(Theorem  4.)  2x   :     2y    :  :  48  :  36 

(Art.  42.)  x    :        y  :  :    4  :    3 

(Theorem  2.)  4:3::  x—y  :     6 

And  4:3::  x+y  :  42 

From  these  last  proportions,  x  —  y=  8 

And     x+y=5Q.     Hence,  x=32,  i/=24. 

2.  Divide    the   number    14   into  two    such   parts,   that    the 
quotient  of  the  greater  divided  by  the  less  shall  be  to  the   quo 
dent  of  the  less  divided  by  the  greater,  as  16  to  9. 

Let  #==  the  greater  part,  and  14  —  a?=the  less. 

x          14  —  x 

By  the  conditions,  -  :   --  :  :  16:9 
14  —  x         x 

Multiplying  terms,       x2  :  (14  —  a?)2  :  :  16:9 
Extracting  root,          x   :  (14—  x)  :  :     4:3     (Theor.  5.; 
Adding  terms,  x    :         14     :  :     4:7 

Dividing  terms,  x   :  2     :  :     4  :  1 

Therefore,  x=S. 

3.  There  are  three  numbers  in  geometrical  progression  whose 
sum  is  52,  and  the  sum  of  the  extremes  is  to  the  mean  as  10  to 
3.     What  are  the  numbers  ?  Ans.     4,  12,  and  36 

Let  x,  xy,  xyz  represent  the  numbers. 
Then,  by  the  conditions,  x-\-xy-\-xyz=52     (I) 

And          xyz-i-x  :  xy  :  :  10  :  3 
(Art.  42.)  yM-1  :  y   :  :  10  :  3 

Double  2d  and  4th,  ?/2+l   :  2y  :  :  10  :  6 

Adding  and  sub.  terms,       £2-f2?/4-l  :  y2—2y-\-l  :  :  16  :  4 
Extracting  square  root,        2/+1  :  y  —  1  :  :  4  :  2 
Adding  and  sub.  terms,        y  :   1  :  :  3  :  1     Hence,  y=3. 

52  52  52 

From  equation  (1),     *=__=__-  =-=4. 


GEOMETRICAL  PROPORTION.  *jll 

4.  The  product  of  two  numbers  is  35,  the  difference  of  their 
cubes,  is  to  the  cube  of   their  difference  as  109  to  4.      What  arc 
the  numbers  ?  Ans.     7  and  5 

Let  x  and  y  represent  the  numbers. 

Then,  by  the  conditions,  xy=35,  and  x2  —  y*  :  (x  —  y)*  :  :  109:4 
Divide  by  (x—y)  (Art.  42.)  and  x2+xy-\-y*  :  (x—y)*  :  :  109:4 
Expanding,  and  #2-f  xy  -\-yz  :  x2  —  2xy-{-yz  :  :  109:4 

(Theorem  3.)  3xy  :  (z-^)?'  :  :  105:4 

ttuiSxy,  we  know  from  the  first  equation,  is  equal  to  105. 
Therefore,  (x  —  y)2=4,  or  x  —  y=2. 

We  can  obtain  a  very  good  solution  of  this  problem  by  putting 
x-\-y=  the  greater,  and  x  —  y=  the  less  of  the  two  numbers. 

5.  What  two  numbers  are  those,  whose  difference  is   to  their 
sum  as  2  to  9,  and  whose  sum  is  to  their  product  as  18  to  77? 

Jlns.     1  1  and  7 

6.  Two  numbers  have  such  a  relation  to  each  other,  that  if  4 
be  added  to  each,  they  will  be  in  proportion  as  3  to  4  ;  and  if  4 
be  subtracted  from  each,  they  will  be  to  each  other  as  1  to  4 
What  are  the  numbers  ?  Ans.     5  and  8 

7.  Divide  the  number  16  into  two  such  parts  that  their  pro- 
duct shall  be  to  the  sum  of  their  squares  as  15  to  34. 

.  10,  and  6. 


8.  In  a  mixture  of  rum  and  brandy,  the  difference  between 
the  quantities  of  each,  is  to  the  quantity  of  brandy,  as  100  is  to 
the  number  of  gallons  of  rum;  and  the  same  difference  is  to  the 
quantity  of  rum,  as  4  to  the  number  of  gallons  of  brandy. 
How  many  gallons  are  there  of  each  ? 

tins.     25  of  rum,  and  5  of  brandy. 

«  9«  There  are  two  numbers  whose  product  is  320  ;  and  the 
difference  of  their  cubes,  is  to  the  cube  of  their  difference,  as  61 
to  1.  What  are  the  numbers?  Am.  20  and  16. 

1O.  Divide  60  into  two  such  parts,  that  their  product  shall  be 
to  the  sum  of  their  squares  as  2  to  5.  Ans.  40  and  20 


21-2  ELEMENTS  OF  ALGEBRA. 

11.  There  are   two  numbers  which  are  to  each  other  as  3  to 
2.     If  6  be  added  to  the  greater  and  subtracted  from  tie  less, 
the  sum    and  the  remainder  will  be  to  each  other,  as  3  to  1. 
What  are  the  numbers  ?  Jlns.     24  and  1 6. 

12.  There  are  two  numbers,  which  are  to  each  other,  as  16 
co  9,  and  24  is  a  mean  proportional  between  them.      What   are 
the  numbers  1  Ans.     32  and  18. 

13.  The  sum  of  two  numbers  is  to  their  difference  as  4  to  1, 
and  the  sum  of  their  squares  is  to  the    greater  as  102  to    5. 
What  are  the  numbers  1  Jlns.     15  and  9. 

14.  If  the  number  20  be  divided  into  two  parts,  which  are  to 
each  other   in  the  duplicate  ratio  of  3  to  1,  what  number  is  a 
mean  proportional  between  those  parts  ? 

Jlns.   18  and  2  are  the  parts,  and  6  is  the  mean  proportion 
between  them. 

15.  There  are  two  numbers  in  proportion  of  3  to  2  ;  and  if 
6  be  added  to  the  greater,  and  subtracted  from  the  less,  the  results 
will  be  as  3  to  1.     What  are  the  numbers  ?      Jins.     24  and  16. 

16.  There  are  three  numbers  in  geometrical  progression,  the 
product  of  the  first  and  second,  is  to  the  product  of  the  second 
and  third,  as  the  first  is  to  twice  the  second ;  and  the  sum  of 
the  first  and  third  is  300.     What  are  the  numbers  ? 

Jlns.     60,  120,  and  240. 

17.  The  sum  of  the  cubes   of  two  numbers,  is  to  the  differ- 
ence of  their  cubes,  as  559  to  127  ;  and  the  square  of  the  first, 
multiplied  by  the  second,  is  equal  to  294.      What  are  the  num- 
bers ?  rfns.      7  and  6. 

18.  There  are  two  numbers,  the  cube  of  the  first  is  to  the 
square  of  the  second,  as  3  to  1 ;  and  the  cube   of  the  second  is 
to  the  square  of  the  first  as  96  to  1.     What  are  the  numbers  ? 

•tins.     12  and  24. 


BINOMIAL  THEOREM.  213 

SECTION    VI. 

CHAPTER  I. 

INVESTIGATION   AND   GENERAL  APPLICATION  OF 
THE  BINOMIAL  THEOREM. 

(Art.  127.)  It  may  seem  natural  to  continue  right  on  to  the 
higher  order  of  equations,  but  in  the  resolution  of  some  cases  in 
cubics,  we  require  the  aid  of  the  binomial  theorem  ;  it  is  there- 
fore requisite  to  investigate  that  subject  now. 

The  just  celebrity  of  this  theorem,  and  its  great  utility  in  the 
higher  branches  of  analysis,  induce  the  author  to  give  a  general 
demonstration  :  and  the  pupil  cannot  be  urged  too  strongly  to 
give  it  special  attention. 

In  (Art.  67.)  we  have  expanded  a  binomial  to  several  powers 
by  actual  multiplication,  and  in  that  case,  derived  a  law  for  form- 
ing exponents  and  coefficients  when  the  power  was  a  whole 
positive  number ;  but  the  great  value  and  importance  of  the 
theorem  arises  from  the  fact  that  the  general  law  drawn  from  that 
case  is  equally  true,  when  the  exponent  is  fractional  or  negative, 
and  therefore  it  enables  us  to  extract  roots,  as  well  as  to  ex- 
pand powers. 

(Art.  128.)  Preparatory  to  our  investigation,  we  must  prove 
the  truth  of  the  following  theorem  : 

If  there  be  two  series  arising  from  different  modes  of  ex- 
panding the  same,  or  equal  quantities,  with  a  varying  quan- 
tity having  regular  powers  in  each  series  ;  then  the  coefficients 
of  the  same  powers  of  the  varying  quantity  in  the  two  series 
are  equal. 

For  example,  suppose 

Jl+Bx+Cy?+nx*,  &c.  =a-\-bx+cxz+dx*,  &c. 
This  equation  is  true  by  hypothesis,  through  all  values  of  a?. 
It  is  true  then,  when  #=0.  Make  this  supposition,  and  <ft=a 
Now  let  these  equal  values  be  taken  away,  and  the  remainder 
divided  by  x.  Then  again,  suppose  #=0,  and  we  shall  find 
fi=b.  In  the  same  manner  we  find  C—c,  D=d,  E=e,  &c. 


214  ELEMENTS  OF  ALGEBRA. 

(Art.  129.)  A  binomial  in  the  form  of  a-\-x  may  be  put  in 
the  form  of  a  X  f  1  -j-  -  )  ;  for  we  have  only  to  perform 
the  multiplication  here  indicated  to  obtain  a-\-x.  Hence 


(x\m 
1  +  -  )  >    it  will  be  sufficient  to  mul- 

tiply every  term  of  the  expanded  series  by  am  for  the  expansion 
of  («-j-.r)OT,  but  as  every  power  or  root  of  1  is  t,  the  first  term 

/        x  \m 
of  the  expansion  of   (  l-f~  —  J     is  1,  and  this  multiplied  by  am 

must  give  am  for  the  first  term  of  the  expansion  of  (a-\-x)m,  what- 
ever m  may  be,  positive  or  negative,  whole  or  fractional. 

99 

As  we  may  put  x  in  place  of  -,   we  perceive  that  any  bino- 

mial may  be  reduced  to  the  form  of  (1-j-o?),  which,  for  greater 
facility,  we  shall  operate  upon. 

(Art.  130.)  Let  it  be  required  to  expand  (l-f-a;)m,  when  m  is 
a  positive  whole  number.  By  actual  multiplication,  it  can  be 
shown,  as  in  (Art.  67.)  that  the  first  term  will  be  1,  and  the 
second  term  mx.  For  if  m=2,  then 

(l+a?)m=(l+a?)8=l+2ar,  &c. 
If  m=3,      (l-hr)ra  =l+3x,  &c. 
And  in  general,         (l-\-x)m=l+mx-\-tfxz,+JBx*,  &c. 

The  exponent  of  x  increasing  by  unity  every  term,  and  .#, 
It,  C,  &c.,  unknown  coefficients,  which  have  some  law  of  de- 
pendence on  the  exponent  m,  which  it  is  the  object  of  this  investi- 
gation to  discover. 

(Art.  131.)  Now  if  m  is  supposed  to  be  a  fraction,  or  if  m=-, 
the  expansion  of  (l-}-x)m  will  be  a  root  in  place  of  a  power,  and 

we  must  expand  (l-\-x)r  .  _i 

For  example,  let  us  suppose  r=2,  then  r 


BINOMIAL  THEOREM.  21  £ 

and  to  examine  the  form  the  series  would  take,  let  us  actually 
undertake  to  extract  the  square  root  of  (l-\-x)  by  the  common 
rule. 

OPERATION. 


2-r-ar  _i^ 

Thus  we  perceive  that  in  case  of  square  root,  the  first  term  of 
the  series  must  be  unity,  and  the  coefficient  of  the  second  term 
is  the  index  of  the  binomial,  and  the  powers  of  x  increase  by 
unity  from  term  to  term. 

We  should  find  the  same  laws  to  govern  the  form  of  the  series, 
if  we  attempted  to  extract  cube,  or  any  other  root;  but,  to  be 
general  and  scientific,  we  must  return  to  the  literal  expression 


Now  as  any  root  of  1  is  1,  the  first  term  of  this  root  must  be 
1,  and  the  second  term  will  have  some  coefficient  to  x.  Let 
that  coefficient  be  represented  by  p  ;  and  as  the  powers  of  r  will 
increase  by  unity  every  term,  we  shall  have 


\     &c. 

Take  the  r  power  of  both  members,  and  we  shall  have 
l+x^^+px+dx2,     &c.)r 

As  r  is  a  whole  number,  we  can  expand  this  second  mem- 
ber by  multiplication  ;  that  is,  by  (Art.  130.),  the  second  mem- 
ber must  take  the  following  form 

l+*=l-|-rjoa:-M'*2,    &c. 

Drop  1,  and  divide  by  x,  and  we  have 


Ky  (Art.  128.)  l=rp      or      p=-  ; 


210  ELEMENTS  OF  ALGEBRA. 

That  is,  the  coefficient  represented  by  p  must  be  equal  to  tho 
index  of  the  binomial. 

Therefore  (\-\-x)r  =  \  +  \.x-{-  Ay?-\-  Bx*  -f,  &c.  ;  the  same 
general  form  as  when  the  exponent  was  considered  a  whole 
number. 

(Art.  132.)  If  we  take  m=-  we  have  to  expand  the  root  of 
a  power.  The  first  term  must  be  1,  and  the  second  term  will 
contain  a?,  with  some  coefficient,  and  the  coefficients  of  x  will 
rise  higher  and  higher  every  term. 

n 

That  is,  (l+x)^  =  l+7?^+^^2,   &c.     Take  the  r  power  of 
both  members,  and  (l-}-x)n=(l-\-pxt  &c.)r. 
Expanding  both  members,  as  in  (Art.  130), 

l+nx-{-axz,  &c.=l+  rp.r-f-.tftf2,  &c. 
Now,  by  (Art.  128),     71=773     or    p=^. 

n 

Therefore,    (1-J-J?)  r=l+  ™x+Jlx*+E3?,  &c.  ;  the  first  two 

terms  following  the  same  law,  relative  to  the  exponents,  as  in  the 
former  cases.     Now  let  us  suppose  m  negative.     Then 

(1-Hr)1*   will  become   (l-\-x)~m=  (Art.  18.) 

Or  by  (Art.  130.)     —  -  *      2 

;     l+maj-f-^a:2,  &c. 


By  actual  division,  l+mx+rfx*,  &c.)  1     (J  —  mx-\-Jl'3f,  &c. 


—  mx  —  £3? 

—  mx  —  m2x2 


That  is,  (\-\-x)~m=l—mxJrJllyr1  which  shows  that  the  same 
general  law  governs  the  coefficient  of  the  second  term,  as  in  the 
former  cases. 

Hence  it  appears  that  whether  the  exponer*  m  of  a  binomial 


BINOMIAL  THEOREM.  217 

be  positive  or  negative,  whole  or  fractional,  th«  same  general 
form  of  expression  must  be  preserved. 

That  is,  in  all  cases  (l+ar)m=l+roar+-foM-  Btf,  &c. 

(Art  133.)  For  clearness  of  conception,  let  the  pupil  bear  in 
mind  that  the  coefficients  of  an  expanded  binomial  quantity 
depend  not  at  all  on  the  magnitudes  of  the  quantities  themselves, 
but  on  the  exponent.  Thus,  (a-\-b)  to  the  5th,  or  to  any  other 
power,  the  coefficients  will  be  the  same,  whether  a  and  b  are 
great  or  small  quantities,  or  whatever  be  their  relation  to  each 
other. 


(Art.  134.)  The  equation  (l+x^^l+mx+rftf+Ba*,  &c., 
is  supposed  to  be  true,  therefore  it  must  be  true,  if  we  square 
both  members.  But  we  have  only  a  portion  of  one  member. 
We  have,  however,  as  much  as  we  please  to  assume,  and  suffi- 
cient to  determine  the  leading  terms  of  its  square,  which  is  all 
that  we  desire.  Square  both  members,  and  (l-}-2a:-{-#8)m= 
(l+mH-^-l-^-f-Ck4,  &c.)2. 

By  expanding  the  second  member,  and  arranging  the  terms 
according  to  the  powers  of  tf,  we  shall  have 


Now  if  we  assume  y—Zx-}-^,  the  first  member  of  this  equa- 
tion becomes  (l-j-t/)m.  If  we  expand  this  binomial  into  a  series. 
it  must  have  the  same  coefficients  as  the  expansion  of  (l-f-#)OT, 
because  the  coefficients  depend  entirely  on  the  exponent  m. 
(Art.  133.) 

Therefore,      l+*=l+rm/-Mi2-r-8-f  &c- 


Put  the  values  of  y,  T/*,  y3,  &c.,  in  this  equation,  and  arrange 
the  terms  according  to  the  powers  of  #,  and  we  have 


SB 


*t  <fcc.     (2) 


The  left  hand  members  of  equations  (1)  and  (2),  are  identical, 
19 


218  ELEMENTS  OF  ALGEBRA. 

and  the  coefficients  of  like  powers  in  the  second  member  must 
be  equal.     (Art.  128.) 

Therefore,  4^-l-m=m2+2^,    or 
And     8 


mu        r  r>  -  -  -  /o\ 

Therefore  B=~±~-  —  ^=m.  —  -—  --  —  -  (3) 

o  .          «  o 

By  putting  the  coefficients  of  the  fourth  powers  of  x  equal, 
we  have 


To  obtain  the  value  of  C  from  this  equation,  in  the  requisite 
form,  is  somewhat  difficult. 

We  must  make  free  use  of  the  preceding  values  of  Jl  and  B, 
which  are  alone  sufficient  ;  but,  to  facilitate  the  operation,  we 
shall  make  use  of  the  following  auxiliary. 

Assume         P=m  —  2,  then  P-\-i—m  —  1, 

m,P-\-m        m  —  1      -  ,  . 

and  o       =m.—-=rf  (a) 

Also,  by  inspecting  equation  (3),  we  perceive  that 

35=^P,        and         2m£=^^          (b) 

3 

By  putting  the  values  of  125  and  2m5  in  the  primitive  equa- 
tion, and  dividing  every  term  by  .#,  and  in  the  second  member 
taking  the  value  of  Jl  from  equation  («),  and  we  shall  have 


_i_^  p,, 
_____  :__      __ 

Multiply  by  6,  and  substitute  the  value  of 
24F-}-  6  =24?7i  —  42,  because  P=m  —  2,  and  we  have 

84O 

__-|-24m—  42=4m/M-3mP-r-3ra. 
Jl 

Collecting  terms,  and  dividing  by  7,  gives 


BINOMINAL   THEOREM.  2-1-9 

But  3m  —  6=3P,  which  substitute  and  transpose,  and 

ll£:=mP_3P=P(m—  3) 
yl 

Or          C^f(m~3)=m  m~1      m~~     m~3 

~~12  2          ~~3~"        4 

Therefore  the  development  of  (l-\-x)m  must  be 


2  2 


234 

whatever  be  the  value  of  m,  positive  or  negative,  whole  or 
fractional,  and  thus  far  the  law  of  development  has  been  demon- 
strated, and  we  infer,  and  can  only  infer,  that  this  law  will 
continue  true,  whatever  be  the  number  of  terms.  Hence,  the 
demonstration  is  complete,  only  so  far  as  we  extend  it. 

This  series  will  terminate  when  (771)  is  a  whole  positive 
uumber,  but  in  all  other  cases  it  will  be  infinite. 

Few  pupils,  who  pay  attention,  find  difficulty  in  compre- 
hending the  preceding  method  of  demonstration,  and  for  that 
reason  we  preferred  it  ;  but  to  the  following  demonstration,  no 
objection  as  to  completeness  or  perfection  can  be  found. 

SECOND   DEMONSTRATION. 
Assume     (l+x)*==l+Ax+£xa  +  Cx*+2)x*-}-  &c.  (1) 

Then         (\Jryy=\JrAy+By*  +  Cy*-\-Dy*  +  &c>  W 
By  subtracting  (2)   from   (1)  and  dividing  the  result  by 
(ff  —  y)  we  obtain 


x—  y  x—y  x—y  x—y 

+  <&c.     (3) 

Place  (l+z)=P,  and(l+y)  =  £.  (a) 

Then  x—y=P—Q.  Whence  (3)  becomes 

^U  c 


—  y 

If  we  divide  (x2  —  y2  )  by  (x  —  y)  the  quotient  will  term- 
inate with  the  second  term,  and  (x3  —  y3  )  divided  by  (x  —  y)  the 
quotient  will  terminate  with  its  third  term,  and  so  on. 


220  ELEMENTS  OF  ALGEEBRA. 

Then  if  n  is  a  whole  number,  the  division  of  the  first  member 
as  indicated  will  terminate  with  the  nth  term  of  the  quotient. 

Now,  divide  m  the    last    equation  as  indicated,  and  after- 
ward suppose  x—y.     It  then  follows  that  P=  Q,  and  we  have 
p*-i_^p*-a  g_|_p»-»  £3_|_  <£c>  to  n  terms.     And  if  P=  Q  the 
whole  quotient  must  be  nPD~  l  for  the  first  member. 

Whence,    nP*-l=A+2Bx-{-3Cxs-{-4I>x3+5Ex*  &c. 

But  P     =  l+x. 

Multiply  these    two    equations,   member    by    member,  and 
we  shall  have, 


But  we  perceive,  by  inspecting  equation  (1),  that 


By  equating  the  right  hand  members  of  these  two  equa- 
tions, and  observing  that  the  coefficients  of  the  like  powers  of 
x,  must  be  equal,  we  shall  have 


or  = 

or         C 

or        D= 
„ 

<KC. 


Whence  A=n 


2 

n—  1  n—2 
_-— 

n  —  1  n  —  2  n  —  3 


234 

and  so  on,  the  law  of  continuation  being  obvious  from  the 
equations,  to  whatever  term  it  may  be  extended. 

Thus  the  demonstration  is  complete  when  n  is  a  whole  posi- 
tive number. 

Now  let  n  be  an  irreducible  fraction,  as  ~, 


BINOMIAL  THEOREM.  221 

1  ,.x 

Then    (l+*)t==: 


And 

j.  i^ 

Now  assume  ( 1  -f-#) l  =P         and       ( 1  +y ) l  = 


8 

8 


Then,       (J+*)1   =P 
And  l+a= 

Whence  z—y=Pi  —  (>*  (c) 

Now  if  we  subtract  (b)  from  («),  and  divide  the  remainder 
by  x  —  y,  we  shall  have 


Now  divide  numerator  and  denominator  of  the  first  member 
by  (P  —  Q),  and  it  will  become 


,  «fcc. 


Now,  because  5  and  J  are  whole  numbers,  the  division  in  both 
numerator  and  denominator  will  be  complete,  and  if  we  sup- 
pose x=y,  it  follows  that  P=  Q,  and  the  results  of  the  several 
divisions  will  be 

+,  &c.  (e) 


But          JP*    =!+«. 


Multiplying    these    equations,    member    by    member,   will 
produce 

,  3 


But  by  inspecting  equation  (a),  and  observing  that  (l-J-#)r 


222  ELEMENTS  OF  ALGEBRA. 


s 


-P  =-+-.Ax+-Bx*-\--Cx*+,  Ac.  (g) 

t  t      t        '  t  t 

Equating  (/)  and  (g),  and  operating  as  before,  we  find 


A=f- 


3        *      2       3~' 

&c.  &c. 

The  same  law  of  development  as  before,  and  this  completes  the 
demonstration  for  all  positive  fractional  exponents. 

Now,  let  the  exponent  be  a  negative  fraction, 

Then,    (l~}-3;)-^=l+Ax-{-£xa-\-Cx3-^,  &c.          (a') 

* 

g  j 

Now  assume,   (l-fz)-7=  />"          Then,  (l+#)-T=  p~4 

Or,  Pt=i_|^. 

In  like  manner,  Q l  = 


By  subtraction,          ^P  l  —  Q  l  —x  —  y 

Subtracting  (V)  from  (a')  and  dividing  by  (x  —  y),  as  in  the 
other  cases,  we  shall  obtain 

Ac. 


The  first  member  of  this  equation  is 
1          1 


»- 


BINOMIAL  THEOREM.  223 


But         _=- 
t 


When  the  division  is  performed,  and  P  supposed  equal  to  Q. 
hence,     —  *  P~8"1  =A+2 

But,  Pi=l+x. 


Whence,     —  *  P~8"1  =A+2£x+3  Cx2  +,  &c. 


Therefore,   — *-P    *  = 


A 


-,  <fec. 


But,  — -  P      =-JL—LAa;-i£x*  — , 

t  t      t  t 


Whence,          A=—8-      J5=—  l.lil!,  &c. 
t  t      2 

and  thus  the  demonstration  is  shown  to  be  complete,  under 
all  possible  circumstances. 

We  may  use  either  of  the  three  following  forms  :* 

Ac.  (1) 
^+,&c.(2) 


2     a2  2        3      a 


2        3        a 

&c.     (3) 
_1  ;!!Z^a»-3;z.3+>  &c. 

»  0 

APPLICATION". 

I.  Required  the  square  root  of  (l-|-a?). 
Apply  formula   (1),  making  w*=-|-  ;    then  the  development 


2.4     2.4.6     2.4.6.8 
The  law  of  the  series,  in  almost  every  case,  will  become 
apparent,  after  expanding  three  or  four  terms,  provided  we 
keep  the  factors  separate. 

*In  practical  examples  we  may  be  required  to  expand  (1  —  x)  as  well  as 
(1-t-tf),  and  the  exponent  may  be  negative  as  well  as  positive.  But  in  all 
cases  the  products  in  the  second  member  must  conform  to  the  rules  of  mul- 
tiplication. 


224  ELEMENTS  OF  ALBKBRA. 

The  above  will  be  the  coefficients  for  any  binomial  square 
root  (Art.  133  );  hence  the  square  root  of  2  is  actually  expressed 
in  the  preceding  series,  if  we  make  a?=l. 

Then  (1-f  1)5==1-K_-L,  &c.  The  square  root  of  3  w 
expressed  by  the  same  series,  when  we  make  #=2,  &c. 

2.  Required  the  cube  root  of  (a-\-b)  or  its  equal,  af  H  —  V 
Formula  (2)  gives 


This  expresses  the  cube  root  of  any  binomial  quantity,  or  any 
quantity  that  we  can  put  into  a  binomial  form,  by  giving  the 
proper  values  to  a  and  b.  For  example,  required  the  cube  root 

of  10,  or  its  equal,  8+2.     Here  a=8,   6=2.     Then   «^=2, 

and  b-=$. 
a 

1  2  25 

Therefore,   2(1+-^—  -—  z+^  Q  'Q  ^  &c.)  is  the  cube  root 

of  10,  and  so  for  any  other  number. 

S.  Expand    -  —  -r  into  a  series,  or  its  equal,   (a  —  b)~l. 

1     b   .b*b* 


b*      64   .     b* 


4.  Expand    (a2-h&*)  into  a  series,  or  its  equal,  «M-|—  2 


Expand   d(c*-}-x?)  *    into  a  series. 
d          a? 


BINOMIAL  THEOREM.  225 

6.  Expand  (a2 — #*)*  into  a  series. 


Ans. 


f 

25a2     27a4 


7.  Expand 


8.  Find  the  cube  root  of 


a3+b 


9.  Find  the  cube  root  of  31. 

31=27+4=27/^1+—^ 

277- 


,  . 

.  1  —  -  +  --  —  -  ,  &c 
3a3      9a6      8  la9 


Place-i=a 


27/  27 


Ans.  3l+_-.—  +.--3+,&c.     =3.156066 
V       327     36  (27)^369  (27)3  / 

1O.  Find  the  seventh  root  of  2245,  or  any  other  number,  by 
the  binomial  theorem. 

N.  B.  Find  by  trial  the  greatest  seventh  power,  less  than  the 
given  number,  (2245),  and  divide  the  given  number  by  it. 

Thus,   37=2187 


2187  2187  2187 

Ans.  sf  1+1.  Jl?  __  A  YJLY+&0.  ^=3.  01  13574 
\     '  7  2187     7.14  \2187/  / 

11.  Find  the  cube  root  of  9,  by  the  binomial  series. 
9=8(1+|).         Cube  root, 


Ans.    2.080084. 
19.'  Find  the  cube  root  of  7,  by  the  binomial  series. 

7=(8—  1)=8(1—  £).          Cube  root,  2(1—  J)* 

Ans.    1.912933. 
13.  Expand  (x—  -y)"1,  into  a  series. 

n—  1 
Ans.  x*  —  tt£B 


226  ELEMENTS  OF  ALGEBRA. 

When  n  =  —  1,  this  series  becomes 

V^JJ/3  ,  y3  , 
SF?Fi»TF^,  &c.,  Ac. 

14.  Find  an  approximate  cube  root  of  100,  by  the  binomial 
series. 

N.  B.  The  nearest  cube  number  to  100,  is  125. 

/        25  \ 
Therefore,  we  place  100=125  —  25=125^1  —  ~\%§). 

Or,     100=125(l—£).     Whence  (100)^=5(1—  $)%. 

Ans.  4.64159,  nearly. 

Thus,  by  a  little  artifice,  this  series  can  be  used  to  extract 
the  cube  (or  any  other  root),  of  any  number. 

15.  Expand  -—  —  into  a  series. 

a+cx 

N.  B.  It  is  much  easier  to  expand  all  such  expressions  as 
this,  by  actual  division,  than  by  the  binomial  theorem,  and 
we  take  the  example,  merely  to  show  that  it  can  be  expanded 
by  the  binomial  theorem. 

The  above  expression  is  the  same  as  1  —  (  ~rc)xt 

a-\-cx 

Place  £-J-c=w,  then  we  are  required  to  expand 


/          M,          nix/*  ,  cx\° 
mx(a+cx)n    or,  _(!+—  ) 
a  \        a/ 


because  n=  —  1. 


Because  n=  —  1,  this  series  becomes 


— IJL-i,  Ac. 
a5 


this  multiplied  by       ,  and  the  product  subtracted  from  1,  pro- 
a 

,  mx  ,  mcx2     me2  x3  .  mc3x*     mc*x5     «        .,  u 

duces    1  —  _  -4-  ___  -  —  --  -t-  ---  —  -  ,  &c.  the   result 
a        a2  a2  a4  a5 

sought,  when  (b-\-c)  is  written  in  place  of  m. 


INFINITE  SERIES.  227 

16.   Expand      '     ,      or  its  equal  14-  _  ,      or  its  equal 
x  —  1  x  —  1 

-^. 
1—  x 

Ans.     —  1—  2*—  2s3—  2*3—  2#4—  ,  &c.,  &o. 
17    Expand  z(l—  a?)-"3  into  a  series. 

Ans.    *+3:r2+6z3-f-10z4-f. 

18.  Expand  (64+1)*,  or  its  equal,  8(1+^)*,  into  a  series. 
N.  B.  This  is  the  same  as  demanding  the  cube  root  of  65. 


CHAPTER  II. 

OF  INFINITE  SERIES. 

(Art  136.)  An  infinite  series  is  a  continued  rank,  or  progres- 
sion of  quantities  in  regular  order,  in  respect  both  to  magnitudes 
and  signs,  and  they  usually  arise  from  the  division  of  one  quantity 
by  another. 

The  roots  of  imperfect  powers,  as  shown  by  the  examples  in 
the  last  article,  produce  one  class  of  infinite  series.  Some  of 
the  examples  under  (Art.  121.)  show  the  geometrical  infinite 
series. 

Examples  in  common  division  may  produce  infinite  series  for 
quotients  ;  or,  in  other  words,  we  may  say  the  division  is  con- 
tinuous. Thus,  10  divided  by  3,  and  carried  out  in  decimals, 
gives  3.3333,  &c.,  without  end,  and  the  sum  of  such  a  series  is 
3|.  (Art.  121.) 

(Art.  137.)  Two  series  may  appear  very  different,  which  arise 
from  the  same  source  ;  thus  1,  divided  by  1-f-a,  gives,  as  we  may 
bee,  by  actual  division,  as  follows: 

l-f-a)l       (1—  a-j-a2  —  a3-f-a4,  <fcc.  without  end. 


228  ELEMENTS  OF  ALGEBRA. 

Also,  «+!)!  ( -+- j,  &c.  without  end. 

'  \  a     a*    a*    a* 


a 

These  two  quotients  appear  very  different,  and  in  respect  to 
blngle  terms  are  so  ;  but  in  these  divisions  there  is  always  a  re- 
mainder, and  either  quotient  is  incomplete  without  the  remain- 
der for  a  numerator  and  the  divisor  for  a  denominator,  and  when 
these  are  taken  into  consideration  the  two  quotients  will  be 
equal. 

We  may  clearly  illustrate  this  by  the  following  example  : — 
Divide  3  by  l-f-2,  the  quotient  is  manifestly  1 ;  but  suppose  them 
literal  quantities,  and  the  division  would  appear  thus  : 

(3— 6+12,  &c. 
3+6 

—6 
—6—12 


12 
12+24 

—24 

Again,  divide  the  same,  having  the  2  stand  first. 

|+f,&c. 


Now  let  us  take  either  quotient,  with  the  real  value   of  its  re- 
mainder, and  we  shall  have  the  same  result. 

Thus,  3+12=15;  and  —6,  and  the  remainder — 24  divided 


INFINITE  SERIES.  229 

by  3,  gives  — 8,  which  make   — 14  ;  hence,  the  whole  quotient 
is  1. 

Again,  H-t=V,  and— |—i=l. 

Hence,  y — |=f =1»  the  proper  quotient. 

If  we  more  closely  examine  the  terms  of  these  quotients,  we 
shall  discover  that  one  is  diverging,  the  other  converging,  and 
by  the  same  ratio  2,  and  in  general  this  is  all  a  series  can  show, 
the  degree  of  convergency. 

(Art.  138.)  We  convert  quantities  into  series  by  extracting 
the  roots  of  imperfect  powers,  as  by  the  binomial,  and  by  actual 
division,  thus : 

1.  Convert   — -. —  into  an  infinite  series. 
a-\-x 

Thus,        «+*«,      (l_+_+,  &c. 


2.  Convert     -  into  an  infinite  series. 
a  —  x 


Observe  that  these  two  examples  are  the  same,  except  the 
signs  of  x  :  when  that  sign  is  plus  the  signs  in  the  series  will  be 
alternately  plus  and  minus  ;  when  minus,  all  will  be  plus. 

3.  What  series  will  arise  from  -  ? 

1  —  x 

Jlns.     1+2*+  2*M-2#3,  &c. 
Observe  that  in  this  case  the  series  commences  with  2x.     The 

unit  is  a  proper  quotient,  and  the  series  arises  alone  from  r  -- 

1  —  z, 

the  remainder  after  the  quotient  1  is  obtained. 


230  ELEMENTS  OF  ALQEBRA. 

4.  What  series  will  arise  from   -r-r — ^  ? 


,,,2        ^4 
_  f___^ 

a      a4 

Observe,  in  this  example,  the  term  x,  in  the  numerator  does 
not  find  a  place  in  the  operation  ;  it  will  be  always  in  the  re- 

mainder :  therefore,—^-;  -  will  give  the  same  series. 

2--z 


5.  What  series  will  arise  from  dividing  1  by  1  —  a-}-  a2,  or 


from 


-  -  -,—  -?         rfns.  1-l-a—  a3—  a4-}-a6+a7—  a9—  a10, 
1  —  a-j-cr 


In  this  example,  observe  that  the  signs  are  not  alternately  plus 
and  minus,  but  two  terms  in  succession  plus,  then  two  minus  ; 
this  arises  from  there  being  two  terms  in  place  of  one  after  the 
minus  sign  in  the  divisor. 

6.     What  series  will  arise  from  y~~? 

•fins.  a-j-ar-r-ar2-|-ar3-l-ar4,  &c. 

Observe  that  this  is  the  regular  geometrical  series,  as  appears 
in  (Art.  118.) 

•7.  What  series  will  arise  from   -  —  -  ? 

rfns.     1-f-l-f-l-fl,  &c 

That  is  -  is  1  repeated  an  infinite  number  of  times,  or  infinity, 
a  re/  U  corresponding  to  observations  under  (Art.  60.) 


SUMMATION  OF  SERIES.  231 

8.  What  series  will  arise  from  the  fraction   — —  ? 


If  «=&,  this  series  will  be   -,  repeated  an  infinite  number  of 

times. 

This  series  can  also  be  expanded  by  the  binomial  theorem, 

for  . j=h(a — 6)  .  This  observation  is  applicable  to  seve- 
ral other  examples. 

(Art.  139.)  A  fraction  of  a  complex   nature,  or  having  com- 

j /£ 

pound  terms,  such  as 2,  may  give  rise  to  an  infinite 

1  —&X—dX 

series,  but  there  will  be  no  obvious  ratio  between  the  terms. 
Some  general  relation,  however,  will  exist  between  any  one  term 
and  several  preceding  terms,  which  is  called  the  scale  of  rela- 
tion, and  such  a  series  is  called  a  recurring  series.  Thus 
the  preceding  fraction,  by  actual  division,  gives  l-}-x-{-5xc*-\r 
ISa^-f-^la^+lSltf5,  &c.,  a  recurring  series,  which,  when  carried 
to  infinity,  will  be  equal  to  the  fraction  from  which  it  is  derived. 

l+2a? 
Expand   — — -^  into  a  series 


CHAPTER  III. 

SUMMATION  OF  SERIES. 

(Art,  140.)  We  have  partially  treated  of  this  subject  in  geo- 
metrical progression,  in  (Art.  121)  ;  the  investigation  is  now 
more  general  and  comprehensive,  and  the  object  in  some  respects 
different.  There  we  required  the  actual  sum  of  a  given  number 
of  terms,  or  the  sum  of  a  converging  infinite  series.  Here  the 
series  may  not  be  in  the  strictest  sense  geometrical,  and  we  may 
not  require  the  sum  of  the  series,  but  what  terms  or  fractional 
quantities  will  produce  a  series  of  a  given  convergent]/. 


232  ELEMENTS  OF  ALGEBRA. 

The  object  then,  is  the  converse  of  the  last  chapter  ;  and  for 
every  geometrical  series,  our  rule  will  be  drawn  from  the  sixth 

example  in  that  chapter  ;  that  is,  -  -  ,   a  being  the  first  term 

of  any  series,  and  r  the  ratio.     We  find  the  ratio  by  dividing 
any  term  by  its  preceding  term. 

Hence,  to  find  what  fraction  may  have  produced  any  geomet- 
rical series,  we  have  the  following  rule: 

RULE.     Divide  the  first  term    of  the  series  by  the  alge- 
braic difference  between  unity  and  the  ratio. 

EXAMPLES. 

1.  What  fraction  will  produce  the  series  2,  4,  8,  16,  &c? 

2 

Here  a=2,  and  r=2  ;  therefore,    -  —  -       Jlns 

1  —  Z 

2.  What  fraction  will  produce  the  series  3  —  9-J-27  —  81,  &c.? 
Here  0=3,  and  r=  —  3;  then  —  r=3, 

Hence,  -fL- 


3.  What  fraction  will  produce  the  series  ^,  T£7,  &c. 
Here  «=TV  and  r=TV;  therefore, 
10         3         31 


4.  What  fraction  will  produce  the  series, 
_+__3,  &c.  ?     [See  example  1,  (Art.  138.)] 


x  1  a 

Here  0=1,  and  r=  -  ;  then  -  =  —  =  —  , 
a  .x     a-i-x 

a 


5.  What  fraction  will  produce  the  series 
&c.  ?     [See  example  3,  (Art.  138.)  and  the  observation  in  con- 
nection.] 


SUMMATION  OF  SERIES.  233 

2# 

Here  a=2x,  r=x\    then   -  -  ,  will   give  all  after  the  first 

2iX       1  l  % 
term:  therefore,   1+-  --  =-  -  ,    »flns. 

,    w.       e                                                   d     db  .  db3  (/6s 
6.  What  fraction  will  produce  the  series   -  H  —  F  --  4  » 

c     er       c*  c* 

&c.  ?  AM. 


What  fraction  will  produce  the  series  7+-  r^—  i  —  rr~  ' 


'b — ax' 
8.  What  fraction  will  produce  the  series 

2ab 


— 3 ,-»  »'     — FT' 

a       a2      a3  a-J-o 

9.  What  fraction  will  produce  the  series 

!-}-«— a3— a44-a6+a7— a9— a10,  &c.  ? 
See  example  5,  (Art.  138.) 

Put  l+a=6;   then   a3-J-a4=a36,  and  ae+«7=a66,   and  the 
scries  becomes  b — a?b-\-a*b — ,  <fec. 

b         1-fa  1 

Hence,  the  fraction  required  is  — = — 3=^-: — 5=- : — 5. 

1-j-a3     1+a3     1 — «+a2 


10.  What  fraction  will  produce  the  series  a?+a^-f-a^,  &c.  ? 

Ana.     —.. 
1 — x 

If  a?s=lf  this  expression  becomes  j— f~o'  a  S7mbol  of  in 
fmity 

11.  What  fraction  will  produce  the  series  1-f-f+JL,  &c.  ? 

1  5 


— 5—3 


Hence,  the  sum  of  this  series,  carried  to  infinity,  is  2£. 
20 


234  ELEMENTS  OF  ALGEBRA. 

In  the  same  manner,  we  may  resolve  every  question  in  (Art. 

tin.) 

12.  What  is  the  sum  oi  the  series,  or  what  fraction  will  pro- 
duce the  series  1  —  a-\-a?  —  «4-f-a6  —  a'-\-cP  —  ,  &c.  ? 

Jlns. 

13.  What  is  the  value  of  the  infinite  series 


RECURRING    SERIES. 

(Art.  141.)  We  have  explained  recurring  series  in  (Art.  139.) 
and  it  is  evident  that  we  cannot  find  their  equivalent  fractions 
by  the  operation  which  belongs  to  the  geometrical  order,  as  no 
common  relation  exists  between  the  single  terms.  The  fraction 

|_l_2<>' 

---  2>  by  actual  division,  gives  the  series  l-|~3#-f-4#M-7a:8 

L^~~3C'     i3C 

-\-llx*  -\-\8x*,  &c.,  without  termination  ;  or,  in  other  words,  the 
division  would  continue  to  infinity. 

Now,  having  a  few  of  these  terms,  it  is  desirable  to  find  a 
method  of  deducing  the  fraction. 

There  is  no  such  thing  as  deducing  the  fraction,  or  in  fact  no 
fraction   could   exist  corresponding  to  the  given    series,  unless 
order  or  a  law  of  dependence  exists  among  the  terms  ;  therefore 
some  order  must  exist,  but  that  order  is  not  apparent. 
Let  the  given  series  be  represented  by 

d+B+C+D+E+F,  &c. 

Two  or  three  of  these  terms  must  be  given,  and  then  each  sue 
oeeding  term  may  depend  on  two  or  three  or  more  of  its  pre- 
ceding terms. 

In  cases  where  the  terms  depend  on  two  preceding  terms  we 
may  have 

C=mBx-}-nJlx* 
J)=mCx-\-nBx* 


RECURRING  SERIES.  236 

In  cases  where  the  terms,  or  law  of  progression  depend  on 
three  preceding  terms  we  may  have 

fi=m 

E=mDx+n 
F=m 
&c.=&c. 

The  reason  of  the  regular  powers  of  x  coming  in  as  factors, 
will  be  perfectly  obvious,  by  inspecting  any  series. 

The  values  of  m,  n  and  r  express  the  unknown  relation,  or  law 
that  governs  the  progression,  and  are  called  the  scale  of  relation 
We  shall  show  how  to  obtain  the  values  of  these  quantities  in  a 
subsequent  article. 

(Art.  142.)  Let  us  suppose  the  series  of  equations  (1),  to  be 
extended  indefinitely,  or,  as  we  may  express  it,  to  infinity,  and 
add  them  together,  representing  the  entire  sum  of  Jl-\-B-\- 
C-\-D,  <fec.,  to  infinity,  by  S  ;  then  the  first  member  of  the 
resulting  equation  must  be  (S  —  J?  —  B),  and  the  other  member 
is  equally  obvious,  giving 

S—  A— 


A+B—mJlx 

Hence,    S=-—l  --  r  (a) 

1  —  mx  —  no? 

In  the  same  manner,  from  equations  (2),  we  may  find 

s_J+B+G-(Jl+B)mx--Jlna*     ,„ 
1  —  mx  —  nx*  —  rx*  *  ' 

(Art.  143.)  The/orw  of  a  series  does  not  depend  on  the 
value  of  x,  and  any  series  is  true  for  all  values  of  x.  Equations 
(1)  then,  will  be  true,  if  we  make  2=1. 

Making  this  supposition,  and  taking  the  first  two  equations  of 
the  series  (1),  we  have 

C^mB+nA  )  (c) 

And  D=mC+nB  \ 

In  these  equations,  ,/?,  B,  (7,  7J,  are  known,  and  m  and  n  un- 
known ;  but  two  unknown  quantities  can  be  determined  from  two 
equations  ;  hence  m  and  n  can  be  determined 


236  ELEMENTS  OF  ALGEBRA. 

When  the  scale  of  relation  depends  upon  three  terms,  wt 
take  three  of  the  equations  (2),  making  x=l,  and  we  determine 
m,  n,  and  r,  as  in  simple  equations. 

EXAMPLES. 
1.  What  fraction  will  produce  the  series 


To   determine   the   scale   of  relation,  we  have   w#=l,  .#=3, 
£=4,  and  D=7.     Then  from  equations  (c),  we  have 
n+3m=4 
3n+4m=7 

These  equations  give  m=l,  and  72=  1. 
Now  to  apply  the  general  equation  (a),  we  have  Jj=\t  23=3x. 

A+B-mAx     1+3*-*       1+2* 
1  hen    o=—  -  5-=-;  --  ?=;  --  at     •««*• 
1  —  mx  —  nx*       1  —  x  —  ar      1  —  x  —  or 

2.  What  fractional  quantity  will  produce  the  series 

l  +  6z+12arM-48z3-f  120;r4,  &c.,  to  infinity? 
Here  .#=1,  B=Qx,  m=l,  n=6. 

Hence,  by  applying  equation  (a),  we  find  -  -  —  j   for  the 

1  ~^x     o*  , 

expression  required. 

3.  What  quantity  will  produce  the  series 

l_L3;r-f  Sa^-fTar'-i-Qtf4,  to  infinity  ?  l-|-ar 


4.  What  quantity  will  produce  the  infinite  series 

1-f  4^+6^+11^+  28^4+63^,  <fec.  ?  in  which  m=2,  n=—  1, 
r=3. 

[Apply  equation  (6).]  ^ns. 

5.  What  fraction  will  produce  the  series 

,  &c.? 


REVERSION  OF  A  SERIES. 
6.   What  fraction  will  produce  the  series 

«,  &c. 


287 


REVERSION    OF   A    SERIES- 

(Art.  B.)  To  revert  a  series  is  to  express  the  value  of  the  un- 
known quantity  in  it,  (which  appears  in  the  several  terms  under 
regular  powers,)  by  means  of  another  series  involving  the  powers 
of  some  other  quantity. 

Thus,  let  x  and  y  represent  two  indeterminate  quantities,  and 
the  value  of  y  be  expressed  by  a  series  involving  the  regular 
powers  of  x.  That  is 

&c. 


To  revert  this,  is  to  obtain  the  simple  value  of  ar,  by  means 
of  another  series  containing  only  the  known  quantities,  a,  b,  c, 
d,  &c.  and  the  powers  of  y. 
To  accomplish  this,asswme 

*,  &c.* 


Substitute  the  assumed  value  of  x  for  x,  x*,  a?8,  <fcc.,  in  the  pro- 
posed series,  and  transposing  y,  we  shall  have 


—  1 


y-\-aB 


+3  cA*B 


&c. 


*  By  examining  previous  authors,  we  have  found  none  that  explain  the 
rationale  of  such  assumptions :  but  these  are  points  on  which  the  learner 
requires  the  greatest  profusion  of  light.  We  explain  thus :  the  term  x  must 
have  some  value,  positive,  negative,  or  zero,  and  the  series  Ay-\-By2-\-  Cy*, 
fcc.,  can  be  positive  and  of  any  magnitude.  It  can  be  negative,  and  of  any 
magnitude,  by  giving  the  coefficients  A,  B  and  C,  negative  values.  It  can  bo 
zero  by  making  ,y=0.  Therefore  it  can  express  the  value  of  or,  whatever  that 
may  be ;  and  because  the  powers  of  y  are  regular,  the  substitution  of  this 
value  of  x  for  the  several  powers  of  x,  in  the  primitive  series,  will  convert 
that  series  into  regular  powers  of  y,  which  was  the  object  to  be  accomplished. 


238 


ELEMENTS  OF  ALGEBRA. 


As  every  term  contains  y,  \ve  can  reduce  the  equation  by 
dividing  by  y ;  and  afterwards,  in  consideration  that  the  equation 
must  be  true  for  all  values  of  y :  making  y=0,  we  shall  have 


I — 1=0         or 


.*-! 

a 


Continuing  the  same  operation  and  mode  of  reasoning  as  in  (Art. 
128.),  and  we  shall  find,  in  succession,  that 


n 

a 


2bz—ac 


a1 

=     &c. 


Substituting  these  values  of  .#,  B,  C,  &c.,  in  the  assumed 
equation  and  we  have  the  value  of  x  in  terms  of  «,  b,  c,  &c., 
and  the  powers  of  y  as  required,  or  a  complete  reversion  of  the 
series. 

EXAMPLES. 

1.  Revert  the  series     i/=x-}-x'2-}-j:3,  &c. 
Here     a=l  6=1      e=l,     &c. 
Assume  x=^.y-f-%2+Oj/3-[-&c. 

Result,    ^=—    +i/3—  i/4+5—  &c. 


ft.  Revert  the  series     x=y  —  J-  -\--  —  •  -  &c. 

234 

Here  a—  I  b=—%     c=|     J=—  |  &c. 

Assume       y^Ax+Btf+Ctf+Dx*-^     &;c.,  and  find 
the  values  of  rf,  B,  C,  &c.,  from  the  formulas  (F), 

Result,       ^+l.+_+.-.&c 


LOGARITHMS. 

3.  Revert  the  series  y=rc+3a?2+5a>3+7a?4+9a;5,  &c. 

Result  x= 


239 


In  case  the  given  series  is  of  the  form  of  x—ay-\-  by*-}-cy5, 
&c.,  the  powers  of  y  varying  by  2,  the  equations  (F]  will  not 
apply,  and  we  must  assume  y=J2x-\-J3x3-\-Cx5,  &c.,  and  sub- 
stitute as  before,  and  we  shall  find 


(0) 


a10 


In  common  cases,  after  the  coefficients,  as  far  as  D,  are  deter- 
mined, the  law  of  continuation  will  become  apparent,  especially 
if  the  factors  are  kept  separate. 


CHAPTER  IV. 
EXPONENTIAL  EQUATIONS  AND  LOGARITHMS, 

(Art.  144.)  We  have  thus  far  used  exponents  only  as  known 
quantities  ;  but  an  exponent,  as  well  as  any  other  quantity,  may 
be  variable  and  unknown,  and  we  may  have  an  equation  in  the 
form  of  a*=b. 

This  is  called  an  exponential  equation,  and  the  value  of  x  can 
only  be  determined  by  successive  approximations,  or  by  making 
use  of  a  table  of  logarithms  already  determined. 

(Art.  145.)  Logarithms  are  exponents.  A  given  constant 
number  may  be  conceived  to  be  raised  to  all  possible  powers,  and 
thus  produce  all  possible  numbers  ;  the  exponents  of  such  pow- 
ers are  logarithms,  each  corresponding  to  the  number  produced. 

Thus,  in  the  equation  a*=6,  x  is  the  logarithm  of  the  number 
b  ;  and  to  every  variation  of  x,  there  will  be  a  corresponding 


240  ELEMENTS  OF  ALGEBRA. 

variation  to  b  •    a   is   constant,  and  is  called  the  base  of  the  sys 
tern,  and  differs  only  in  different  systems. 

The  constant  a  cannot  be  1,  for  every  power  of  1  is  1,  and  the 
variation  of  x  in  that  case  would  give  no  variation  to  b  ;  hence, 
the  base  of  a  system  cannot  be  unity  ;  in  the  common  system  it 
is  10. 

In  the  equation  10Z=2,  x  is,  in  value,  a  small  fraction,  and 
is  the  logarithm  of  the  abstract  number  2. 

In  the  equation  ax=b,  if  we  suppose  x=l,  the  equation  becomes 
al=a;  that  is,  the  logarithm  of  the  base  of  any  system  is  unity. 

If  we  suppose  37=0,  the  equation  becomes  a°=l  ;  hence,  the 
logarithm  of  1  is  0,  in  every  system  of  logarithms. 

(Art.  146.)  The  logarithms  of  two  or  more  numbers  added 
together  give  the  logarithm  of  the  product  of  those  numbers, 
and  conversely  the  difference  of  two  logarithms  gives  the  lo- 
garithm of  the  quotient  of  one  number  divided  by  the  other. 

For  we  may  have  the  equations  «z=&,  av=b',  and  a*=b". 

Multiply  these  equations  together,  and  as  we  multiply  powers 
ly  adding  the  exponents  the  product  will  be 


Hence,  by  the  definition  of  logarithms,  x-\-y-\-z  is  the  loga- 
rithm for  the  number  represented  by  the  product  bb'b".     Again. 

divide  the  first  equation  by  the  second,  and  we  have  ax~y=j-i; 

and  from  these  results  we  find  that  by  means  of  a  table  of  loga- 
rithms multiplication  may  be  practically  performed  by  addition, 
and  division  by  subtraction,  and  in  this  consists  the  great  utility 
of  logarithms. 

(Art.  147.)  In  the  equation  ax=b,  take  a=10,  and  x  succes- 
sively equal  to  0,  1,  2,  3,  4,  &c. 

Then  10°=1,  10^10,  lO^lOO,  103=1000,  &c. 
Therefore,  for  the  numbers 

1,  10,  100,  1000,  10000,  100000,  &c.,  we  have  for  corres- 
ponding logarithms 

0,  1,  2,  3,  4,  5,  <fcc. 


LOGARITHMS.  241 

Here  it  may  be  observed  that  the  numbers  increase  in  geometri- 
cal progression,  and  their  logarithms  in  arithmetical  progression. 

Hence  the  number  which  is  the  geometrical  mean  between  two 
given  numbers  must  have  the  arithmetical  mean  of  their  lo- 
garithms, for  its  logarithm. 

On  this  principle  we  may  approximate  to  the  logarithm  of  any 
proposed  number.  For  example,  we  propose  to  find  the  /a- 
garithm  of  2. 

This  number  is  between  1  and  10,  and  the  geometrical  mean 
between  these  two  numbers,  (Art.  122.),  is  3.16227766.  The 
arithmetical  mean  between  0  and  1  is  0.5;  therefore,  the  number 
3.16227766  has  0.5  for  its  logarithm. 

Now  the  proposed  number  2  is  between  1  and  3.162,  &c.,and 
the  geometrical  mean  between  these  two  numbers  is  1.778279, 
and  the  arithmetical  mean  between  0.  and  0.5  is  0.25  ;  therefore, 
the  logarithm  of  1.778279  is  0.25. 

Now  the  proposed  number  2  lies  between  1.778279,  and 
3.16227766,  and  the  geometrical  mean  between  them  will  fall 
near  2,  a  little  over,  and  its  logarithm  will  be  0.375.  Continuing 
the  approximations,  we  shall  at  length  find  the  logarithm  of  2  to 
be  0.301030,  and  in  the  same  manner  we  may  approximate  to 
the  logarithm  of  any  other  number,  but  the  operation  would  be 
very  tedious. 

(Art.  148.)  We  may  take  a  reverse  operation,  and  propose  a 
logarithm  to  find  its  corresponding  number;  thus,  in  the  general 
equation  az=n;  x  may  be  assumed,  and  the  corresponding  value 
of  n  computed. 

Thus  suppose  ar=T3ff  ;  then  (10)T»=n,  or  103=n10. 


Hence,  n=™J  1000=1. 9952623 15. 

That  is,  the  number  1.9952,  <fcc.,  (nearly  2)  has  0.3  for  its 
logarithm.  In  the  same  way  we  may  compute  the  numbers  cor- 
responding to  the  logarithms  0.4,  0.5,  0.6,  0.7,  &c. 

(Art.  149.)  We  may  take  another  method  of  operation  to  find 
the  logarithm  of  a  number. 
21 


242  ELEMENTS  OF  ALGEBRA. 

Let  the  logarithm  of  3  be  required. 
The  equation  is  10Z=3,  the  object  is  to  find  x. 
It  is  obvious  that  x  must  be  a  fraction,  for  101==10  ;  there 
fore, 

Let  #=— .     Then  10Z'=3,  or  10=3*.     Here,  we  perceive, 
that  x'  must  be  a  little  more  than  2.     Ma.te  x'=2-{-— p    Then 

Hit  x»      1 0 

32X3    =10;  or  3    =— . 

«7 

Or     (  ~)    =3. 


Here  we  find  by  trial  that  x"  is  between  10  and  1 1 ;  take  it  10, 
and  a?'=2-f-y^ ;  hence,  a?=!,j=0. 476,  for  a  rough  approxima- 
tion for  the  logarithm  of  3.  A  more  exact  computation  gives 
0.4771213;  but  all  these  operations  are  exceedingly  tedious,  and 
to  avoid  them,  mathematicians  have  devised  a  more  expeditious 
method  by  means  of  a  converging  series  ;  which  we  shall  inves- 
tigate in  a  subsequent  article. 

(Art.  150.)  It  is  only  necessary  to  calculate  directly  the  lo- 
garithms of  prime  numbers,  as  the  logarithms  of  all  others  may 
be  derived  from  these.  Thus,  if  we  would  find  the  logarithm 
of  4,  we  have  only  to  double  that  of  2;  for,  taking  the  equation 
(10)z=2,  and  squaring  both  members  we  have,  (10)2z=4 ;  or 
taking  the  same  equation,  and  cubing  both  members,  we  have 
(10)3z=8  ;  which  shows  that  twice  the  logarithm  of  2  is  the 
logarithm  of  4,  and  three  times  the  logarithm  of  2  is  the  loga- 
rithm of  8 ;  and,  in  short,  the  sum  of  two  or  more  logarithms 
corresponds  to  the  logarithm  of  the  product  of  their  numbers, 
(Art.  142.),  and  the  difference  of  two  logarithms  corresponds  to 
the  logarithm  of  the  quotient,  which  is  produced  from  dividing 
one  number  by  the  other. 

Thus,  the  logarithm  of  j  —  log.  a  —  log.  b.  Tne  abbrevia- 
tion, (log.  a),  is  the  symbol  for  the  logarithm  of  a. 


LOGARITHMS.  243 

(Art.  151.)  As  the  logarithm  of  1  is  0,  10  is  1,  100  is  2,  &c., 
we  may  observe  that  the  whole  number  in  the  logarithm  is  one 
less  than  the  number  of  places  in  the  number. 

The  whole  number  in  a  logarithm  is  called  its  characteristic, 
and  is  not  given  in  the  tables,  as  it  is  easily  supplied.  For  ex- 
ample, the  integral  part  of  the  logarithm  of  the  number  67430 
must  be  4,  as  the  number  has  5  places.  The  same  figures  will 
have  the  same  decimal  part  for  the  logarithm  when  a  portion  of 
them  become  decimal. 

Thus,     67430   logarithm  4.82885 

6743.0  3.82885 

674.30  2.82885 

67.430  1.82885 

6.7430  0.82885 

.67430     —1.82885 

For  every  division  by  10  of  the  number,  we  must  diminish 
the  characteristic  of  the  logarithm  by  unity. 

The  decimal  part  of  a  logarithm  is  always  positive  ;  the  index 
or  characteristic  becomes  negative,  when  the  number  becomes 
less  than  unity. 

By  reference  to  (Art.  18.),  we  find  that  — =10-',  — =— = 

10~2,  &c.     That  is  :  fractions  may  be   considered  as  numbers 
with  negative  exponents,  and  logarithms  are  exponents ;  there- 
fore the  logarithm  of  — ,  or  .1,  is  — 1 ;  of  — — ,  or  .01,  is  — 2, 
10  100 

&c.  If  any  addition  is  made  to  .01  the  logarithm  must  be  more 
than  — 2 ;  but,  for  convenience,  we  still  let  the  index  remain 
—2,  and  make  the  decimal  part  plus.  Hence  the  index  alone 
must  be  considered  as  minus. 

Negative  numbers  have  no  logarithms;  for,  in  fact,  as  tot 
have  before  observed,  there  are  no  such  numbers. 

(Art.  C.)  We  now  design  to  show  a  practical  method  of  com- 
puting logarithms.  The  methods  hitherto  touched  upon  were 
only  designed  to  be  explanatory  of  the  nature  of  logarithms ;  but, 
to  calculate  a  table,  by  either  of  those  methods,  would  exhaust  the 


244  ELEMENTS  OF  ALGEBRA. 

patience  of  the  most  indefatigable.  To  arrive  at  easy  practical 
results  requires  the  clearest  theoretical  knowledge,  and  we  must 
therefore  frequently  call  the  attention  of  students  to  first  prin- 
ciples. 

The  fundamental  equation  is  a*=b,  in  which  a  is  the  con 
slant  base  and  x  is  the  logarithm  of  the  number  b ;  and  b  may 
be  of  any  magnitude  or  in  any  form,  if  it  be  essentially  positive. 

Now  we  may  take  a=l+c     and     6=1  -j — 

Then  the  fundamental  equation  becomes  (l-f-c)z=(l-|—  J  (1) 

Here  ar=log.  (l+i)=log.(^i)=log.(/i+l)-]og./i  (2) 

Raise  both  members  of  equation  (1)  to  the  nth  power,  then 
we  shall  have 


Expand  both  members  into  a  series  by  the  Binomial  Theorem, 

Then      l+nxc+nx.  ^^c^+nx .  ^^  .  ^=^c34- 

£  I  • 

nx — 1   nx — 2  nx — 3  4  1  .      n — 1  1  . 

nX— 2—  8— 4-'+'  &C"    =       l+n-j+n-2-p*+ 

n — 1   n — 2    1   ,       n — 1  n— 2  n — 3    1 
n. 


a 3 

Drop  unity,  from  both  members,  and  divide  by  n,  then  we  have 

<,  nx — 1  ,  .  nx — 1     nx — 2  „  .  nx — 1     nx — 2    nx — 3  . 
c+-a-c'+-—  .  ^-c'+-2-  .  -jr-  •  —f-S 

\  In — 1   1   ,  n — 1  n — 2  I     n— 1  n—2  n — 3  1 

"*"       )       ~~  p        2""'/"1      2~'~3~y~h~2~'~3~~T~'^ 

+,  &c. 

This  equation  is  true  for  all  values  of  n.     It  is,  therefore,  true 


LOGARITHMS.  945 

when  »=0.     Making  this  supposition  and  the  above  equation 
reduces  to 

c2     c3     c4     c5  \      1       1,1        1.1 

-2-  +rr  +5—  ^rrw^w^W'^'  w 

As  c  remains  a  constant  quantity  for  all  variations  of  x  and  /), 
the  series  in  the  vinculum  may  be  represented  by  the  symbol  AI 

Then  xM=-  --  -2-f  -L  -  L+J_       &c.         (3) 
jo     2/>2     3/     4jo4     5/ 

Take  the  value  of  x  from  equation  (2),  and  substitute  it  in 
equation  (3), 

Then  [\og.(p+l)-los.p-]M=l~;L+;L-±+,  &c.  (4) 


Again,   we   may  have   the   fundamental   equation 
(  I  -  Y  in  which  y  is  the  logarithm  of  (  I  --  Y  the  same  as 

of(l+l). 


Was 


Or    2/=log.(- )=log.(j9 — 1) — l°g«  P"         (5) 

Operating  on  the  equation   (l-j-c)1/=l ,    the   same   as   w* 

did  on  equation  (1),  we  shall  find 

Subtract  equation  (6)  from  equation  (4),  and  we  obtain 

/I       1         1         1  \ 

[1og.(;;-fl) — log.(p — 1)]M=2( — ^~5~i~^"K~iH ^»  &Ct  )    P; 

Dividing  by  M,  and  considering  that  log/^-  -  )=log.(p-f-l).  • 
log. ( p — 1),  the  equation  can  take  this  form 


246  ELEMENTS  OF  ALGEBRA. 


As  p   may  be  any  positive  number,  greater   than    1,  make 
"  —  -=1-J  —  .     Then  j0=2*4-l,  and  equation  (8)  becomes 


log\~~FL)= 


+ 


3(2H-1)*  '  5(2z-r-l)5' 

By  this  last  equation  we  perceive  that  the  logarithm  of  (z-f-1) 
will  become  known  when  the  log.  of  z  is  known,  and  some 
value  assigned  for  the  constant  M.  Baron  Napier,  the  first  dis- 
coverer of  logarithms,  gave  M  the  arbitrary  value  of  unity,  for 
the  sake  of  convenience. 

Then,  as  in  every  system  of  logarithms,  the  logarithm  of  1  is 
0,  make  z=l,  in  equation  (9),  and  we  shall  have 

h^Lrh  &cA=.O.C9314718. 


)= 


This  is  called  the  Napierian  logarithm  of  2,  because  its  magni- 
tude depends  on  Napier's  base,  or  on  the  particular  value  of  M 
being  unity. 

Having  now  the  Napierian  logarithm  of  2,  equation  (9)  will 
give  us  that  of  3.  Double  the  log.  of  2  will  give  the  logarithm 
of  4.  Then,  with  the  log.  of  4,  equation  (9)  will  give  the  loga- 
rithm of  5,  and  the  log.  of  5  added  to  the  log.  of  2,  will  give  the 
logarithm  of  10. 

Thus  the  Napierian  logarithm  of  10  has  been  found  to  great 
exactness,  and  is  2.302585093. 

The  Napierian  logarithms  are  not  convenient  for  arithmetical 
computation,  and  Mr.  Briggs  converted  them  into  the  common 
logarithms,  of  which  the  base  is  made  equal  to  10. 

To  convert  logarithms  from  one  system  into  another,  we 
may  proceed  as  follows  : 

Let  e  represent  the  Napierian  base,  and  a  the  base  of  the 
common  system,  and  N  any  number. 


LOGARITHMS.  247 

Also,  let  3:  represent  the  logarithm  of  JV,  corresponding  to 
the  base  «,  and  y  the  logarithm  of  N9  corresponding  to  the 
base  c. 

Then         ax=Nt 

and         &  ~N. 

Now,  by  inspecting  these  equations,  it  is  apparent  that  if  the 
base  a  is  greater  than  the  base  e,  the  log.  x  will  be  less  than  the 
log.  y. 

These  equations  give       aT—ey. 

Taking  the  logarithms  of  both  members,  observing  that  x  and 
y  are  logarithms  already,  we  have 

a?  log.  a=y  \og.e. 

This  equation  is  true,  whether  we  consider  the  logarithms 
taken  on  the  one  base  or  on  the  other.  Conceive  them  taken  on 
the  common  base,  then 

log.  0=1,      and          x=y\Q<r*e (10) 

x 
or      log.e=-. 

y 

In  this  equation  x  and  y  must  be  logarithms  of  the  same  num- 
ber, and  therefore  if  we  take  a?=l,  which  is  the  logarithm  of 
10,  in  the  common  system,  y  must  be  2.302585093,  as  pre- 
viously determined. 

Hence  log.f=2—L-— =0.434394482.  .  .  .    (11) 

This  last  decimal  is  called  the  modulus  of  the  common  sys- 
tem ;  for  by  equation  (10)  we  perceive  that  it  is  the  constant 
multiplier  to  convert  Napierian  or  hyperbolic  logarithms  into 
common  logarithms. 

But  equation  (9)  gives  Napierian  logarithms  when  M=\\ 
therefore  the  same  equation  will  give  the  common  logarithms  by 
causing  M  to  disappear,  and  putting  in  this  decimal  as  a  factor. 

Equation  (9)  becomes  the  following  formula  for  computing 
common  logarithms  : 


ELEMENTS  OF  ALG'KBKA 
log.(z-r-l)— •  log.*= 

0.868588961^ 


(-L_+—  ^+^-1--,+,  Ac.)  (F) 


To  apply  this  formula,  assume  z=  10.     Then 
log.  *=1,         and         2z-f-l=21. 


21 


441 


0.86858896 

0.041361379—1   =  0.041361379 

93792 — 3  =  31264 

212—5  =  42 


.041392685=sum  of  series. 
log.lO=    1.0 

log.(*+l)  =    1.041 392685 =log.ll. 

If  we  make   2=99,  then    (z-H)  =  100,  and   Iog.(*-H)=s2, 
and  224-1=199. 


199 
39601 


0.86858896 

436477-H =0.00436477 
11-A_3=  4 


0.00436481  =sum  of  series. 

Hence         2.00000—log.99-=0.00436481 
By  transposition          log.99=l. 99563519 

Subtract  log.l  l=1.04139269=log.l  1 

log.  9=0.95424234 
£log.9=!og.  3=0.47712 117=log.3. 

Thus  we  may  compute  logarithms  with  great  accuracy  ami 
rapidity,  using  the  formula  for  the  prime  numbers  only. 

By  equation  (11)  we  perceive  that  the  logarithm  of  the  Na- 
pierian base  is  0.434294482  ;  and  this  logarithm  corresponds  to 
the  number  2.7182818,  which  must  be  the  base  itself. 

We  may  also  determine  this  base  directly : 

In  the  fundamental  equation  (1),  the  base  is  represented  by 
(1+c).  In  equation  (A),  c  must  be  taken  of  such  a  value  as 


LOGARITHMS. 


249 


c8      c3     c4 
shall  make  the  series  c—  —  +-  —  --f,  &c.,  equal  to  1.     But  to 

determine  what  that  value  shall  be,  in  the  first  place,  put 
c2  ,  c3     c4 

^-sf+s-T  &c' 

Now  by  reverting  the  series  (Art.  B.),  we  find  that 


But,  by  hypothesis,  the  series  involving  c  equals  unity   that  is, 
y=l.     Therefore 


By  taking  12  terms  of  this  series,  we  find  (l-j-c)=2.7182818, 
the  same  as  before. 

TABLE  I.—  LOGARITHMS  FROM  J   TO  100. 


N. 

Log. 

N. 

Log. 

N. 

Log. 

N. 

Log. 

1 

0  000000 

26 

1  414973 

51 

1  707570 

76 

1  880814 

2 

0  301030 

27 

1  431364 

62 

1  716003 

77 

1  886491 

3 

0  477121 

28 

1  447158 

53 

1  724276 

78 

1  892095 

4 

0  602060 

29 

1  462398 

54 

1  732394 

79 

1  897627 

5 

0  698970 

30 

1  477121 

55 

1  740363 

80 

1  903090 

6 

0  778151 

31 

491362 

56 

1  748188 

81 

1  908485 

7 

0  845098 

32 

505150 

57 

1  755875 

82 

1  913814 

8 

0  903090 

33 

518514 

58 

1  763428 

83 

1  919078 

9 

0  954243 

34 

531479 

59 

1  770852 

84 

1  924279 

10 

1  000000 

35 

544068 

60 

1  778151 

85 

1  929419 

11 

1  041393 

36 

556303 

61 

1  785330 

86 

1  934498 

12 

1  079181 

37 

568202 

62 

1  792392 

87 

1  939519 

13 

113943 

38 

579784 

63 

1  799341 

88 

1  944483 

14 

146128 

39 

591065 

64 

1  806180 

89 

1  949390 

15 

176091 

40 

602060 

65 

1  812913 

90 

1  954243 

16 

204120 

41 

612784 

66 

1  819544 

91 

1  959041 

17 

230449 

42 

623249 

67 

1  826075 

92 

1  963788 

18 

255273 

43 

633468 

68 

1  832509 

93 

1  968483 

19 

278754 

44 

643453 

69 

1  838849 

94 

1  973128 

20 

301030 

45 

653213 

70 

1  845098 

95 

1  977724 

21 

322219 

46 

662578 

71 

1  851258 

96 

1  982271 

22 

342423 

47 

672098 

72 

1  857333 

97 

1  986772 

23 

361728 

48 

681241 

73 

1  863323 

98 

1  991226 

24 

380211 

49 

690196 

74 

1  869232 

99 

1  995635 

25 

397940 

50 

698970 

75 

1  875061 

100 

2  000000 

250 


ELEMENTS  OF  ALGEBRA. 


TABLE  II. — LOGARITHMS  OF  LEADING  NUMBERS  WITHOUT  INDICES. 


0 

1 

2 
3 
4 
6 
6 
7 
8 
9 

N.  10Q.  N.  101. 

N.  1(W  !  N.  10^.  N.  104. 

N.  105. 

N.  U)6. 

N.  107. 

N.  108.!  N.  109. 

000000 
000434 
000867 
001301 
001734 
002166 
002598 
003029 
003461 
003891 

0043-21 
004750 
005881 
006609 
006038 
006466 
006894 
007321 
007748 
008174 

003600 
009026 
009451 
009876 
010300 
010724 
011147 
011570 
011993 
012415 

012837 
013259 
013680 
014100 
014521 
014940 
015360 
015779 
016197 
016616 

017033 
017451 
017868 
018284 
018700 
019116 
019532 
019947 
020361 
020776 

021189 
021603 
022016 
022428 
022841 
023252 
023664 
024075 
024486 
024896 

025306 
025715 
026125 
026653 
026942 
027350 
027757 
028164 
028571 
J028978 

029384 
029789 
030195 
030600 
031004 
031408 
031812 
032216 
032619 
1033021 

033424 
033826 
034227 
034628 
035029 
035430 
035830 
036230 
036629 
037028 

0374-26 
037825 
038223 
038620 
039017 
039414 
039811 
040207 
040602 
040998 

(Art.  15la.)  The  preceding  tables  are  sufficient  to  give  us 
the  logarithms  of  any  number  whatever,  provided  we  under- 
stand the  theory  of  logarithms,  and  are  able  to  use  a  few  prac- 
tical artifices. 

If  to  this  knowledge,  and  these  artifices,  we  add  the  formula, 
we  can  with  ease  find  the  log.  of  any  number,  however  large, 
or  however  small,  to  any  required  degree  of  accuracy.* 

We  can,  in  fact,  find  the  logarithm  of  any  number,  very 
nearly,  by  Table  I. 

ILLUSTRATIONS. 

1.  We  require  the  logarithm  of  520;  but  520  is  not  in  the 
table,  52  and  10  are. 


Log.  of  52     is 
Log.  of  10     is 

Log.  of  520,  therefore,  is 

2     The  logarithm  of  5.2  is, 
And  of  .52  is 


1.716003. 
1.000000. 

2.716003. 

0.716003, 
—1.716003,  (fee.,  <fec. 


3.      What  is  the  logarithm  of  146  ? 

146=73X2.     Whence  log.  73,         1.863323 
log.    2,  301030 

Log  of  146  is        2.164353 

*It  is  very  bad  policy  to  give  a  learner  the  use  of  voluminous  tables  ; 
such  tables  are  made  to  supply  the  place  of  theory,  and  of  thought.  Con- 
tracted tables,  like  these,  are  the  best  educators. 


LOGARITHMS.  251 

4.      What  is  the  logarithm  of  3244  ? 

3240=60  •  54.  3250=65  •  50. 

Log.  60,          1.778151  Log.  65,          1.812913 

Log.  54,          1.732394  Log.  50,          1.698970 

Log.  3240,  is    3.510545.  Log.  3250,  is     3.511883. 

(Art.  1515.)  Here  we  have  the  logarithm  of  3240,  and  of 
3250  ;  and  the  number  3244  is  intermediate,  therefore  its  loga- 
rithm must  be  intermediate  between  3.510545  and  3.511883. 
The  difference  between  the  two  extreme  numbers  is  10,  and 
the  difference  of  their  logarithms  is  .001338. 

Therefore,  by  proportion,  10  :  .001338  :  :  4. 

That  is  -^ths  of  the  difference  of  the  known  logarithms  is 
the  correction  to  add  to  the  less  logarithm  to  obtain  the  loga- 
rithm sought. 

Whence  to  log.  3240  3.510545 

Add  .0001338X4  635 

Log.  of  3244  is          3.511080 
This  example  shows  that  Table  I.,  and  the  true  theory  of 

logarithms,  are  sufficient  to  find  the  logarithm  of  any  number 

whatever. 

ANOTHER  METHOD. 
(Art.  151c.)  When  the  number  is  large,  as  in  the  present 

example,  the  first  term  of  the  series  will  be  the  increase  of  the 

logarithm  corresponding  to  one  ;  four  times  this  would  be  the 

increase  corresponding  to  four,  and  so  on. 

The  logarithm  of  3240  is  3.510545,  as  before  found,  and 

here  z=3240,  and  22+1=6481, 

.86858896 
Whence,  —  T-  -  =0.000134 


Whence  to  losr.  3.540545 


Add  0.000134X4  536 


Log.  3244  3.511081 

Ar.d  the  two  methods  agree — the  first  may  be  clearest  to  a 
learner,  but  the  second  is  more  brief  and  scientific. 


252  ELEMENTS  OF  ALGEBRA. 

(Art.  151d.)  Table  II.  requires  some  explanation.  The 
numbers  are  at  the  top  and  side,  and  only  the  decimal  part  of 
the  logarithm  is  contained  in  the  table.  The  operator,  in  each 
particular  case,  will  supply  the  index,  according  to  the  theory 

of  logarithms. 

EXAMPLES. 

1.  Find  the  logarithm  of  1013.  Ans.  3.005609. 
We  find  101  at  the  top,  and  3  at  the  side.    Under  the  former, 

and  opposite  the  latter,  we  find  the  decimal  .005609,  and  this 
is  the  decimal  part  of  the  logarithms  of  all  numbers  expressed 
by  the  figures  1013,  as  they  here  stand. 

Thus,  it  is  the  decimal  logarithm  of  1013, 101.3,  10.13,  1.013, 
or  of  the  decimals  .1013,  .01013,  &c. 

It  is  also  the  decimal  logarithm  of  10130,  101300,  <fcc.,  but 
this  was  sufficiently  illustrated  in  (Art.  151). 

2.  Find  the  logarithm  of  1 .075.  Ans.  0.031408. 

3.  Find  the  logarithm  of  10940.  Ans.  4.039017. 
Here,  under  109,  at  the  top,  and  opposite  4,  at  the  side,  we 

find  the  decimal,  and  we  take  4  for  the  index,  because  there 
are  five  places  of  figures,  in  whole  numbers. 

4.  Find  the  logarithm  of  101 .5.  Ans.  2.031408. 

5.  Find  the  logarithm  o/. 001043.  Ans.  3.018284. 

(Art.  1510.)  Table  II.  will  be  very  useful  in  solving  prob- 
lems in  compound  interest,  and  annuities ;  but  its  greatest 
utility  is  in  solving  the  following  problems  : 

6.  Find  the  logarithm  of  5&1521.  Ans.  5.753977. 
Divide  by  the  two  superior  figures  (in  this  example  it  is  56), 

and  the  quotient  is  10134^|. 

The  divisor,  consisting  of  two  figures,  its  logarithm  can 
always  be  found  in  Table  I. 

And  the  four  superior  figures  of  the  quotient  (any  quotient) 
will  be  found  in  Table  II. 

The  quotient  in  this  example  may  be  written  thus, 
10130+41!-. 

The  logarithm  of  10130  (the  decimal  part  of  it),  can  be 
taken  directly  out  of  Table  II.  It  is  .005609 ;  and  the  next 


LOGARITHMS.  253 

greater  (the  number  below)  is  .006038 ;  the  difference  between 
the  two  is  .000429,  and  the  correction  for  4\%  must  be  4££ 
times  one  tenth  of  .000429,  which  is  .000180. 

Whence  log.  10130    -  4.005609     Table  II. 

Correction  for  4}|,      -         -       .000180 

Log.  of  quotient,  10134^,  4.005789 

Log.  of  56,  -  -     1.748188     Table  I. 

Log.  of  567521,  5.753977. 

Thus  we  may  find  the  logarithm  of  any  number  by  the  use  of 
these  two  tables.     We  give  one  more  example. 

7.     Find  the  logarithm  of  365. 25638 — the  days  and  parts  of 
a  day  in  a  siderial  year. 

(N.  B.  As  three  figures  are  whole  numbers,  the  index  must 
be  two.     Divide  by  36,  and  taking  five  figures  in  the  quotient 
as  whole  numbers,  we  shall  have  10146.01  for  a  quotient.) 
This  quotient  may  be  written  in  the  following  form : 

10140+6.01. 

The  Log.  of  36,  Table  I.,  -     0.556303 

Log.  of  10140,  Table  II.,        -         0.006038 

6.01 
Correction  for  -y^-  of  diff.*  .000257 

Sum  +2.=log.  of  365.25638,         2.562598 

(Art.  151/.)  To  find  the  number  corresponding  to  a  given 
logarithm  we  take  the  converse  operation. 

For  example,  we  take  the  logarithm  just  found,  and  demand 
its  corresponding  number. 

Log.  (omit  index),  0.562598 

Next  less  in  Table  I.       36,        -        .556303 

.006295 
Next  less  in  Table  II.      10140,          .006038 


.000257 


*The  difference  in  the  table  is  the  difference  between  .006038  and  the 
next  greater  .006466,  which  is  .000428,  one-tenth  of  this  .000042.8,  multi- 
ply by  (6.01)  and  the  product  is  .000257. 


254  ELEMENTS  OF  ALGEBRA. 

Divide  this  difference  by  the  tabular  difference,  which  is 
.000428,  and  we  obtain  6,  and  a  very  small  remainder. 

Whence,  to  10140,  add  6,  and  we  have  10146,  which  being 
multiplied  by  36,  produces  365.256,  for  the  number  sought. 

We  take  three  of  the  figures  for  whole  numbers,  because 
the  index  of  the  given  logarithm  is  2.  Had  it  been  3,  we 
should  have  taken  four  figures  for  whole  numbers. 

1.  What  number  corresponds  to  the  logarithm  0.317879  ? 

Ans.  2.079118. 

From  the  given  log.  0.317879 

Sub.  next  less.     Table  I.,     2,         0.301030 

0.016849 
Next  less.     Table  II.,     1039,         0.016616 

0.000233 

Divide  233  by  417  (tab.  diff.),  and  we  obtain  559,  nearly, 
which  we  annex  to  1039,  making  1039559,  which,  multiplied 
by  2,  produces  2.079118,  the  answer. 

2.  What  number  corresponds  to  the  logarithm  4.031718? 

Ans.   10757.673. 

As  the  first  figure  in  the  decimal  is  0,  we  had  better  use 
Table  II.  at  once. 

The  number  next  less  in  the  table  is  the  log.  of  1075.     As 
the  index  is  4,  we  must  have  another  figure  for  whole  numbers. 
Hence,  from  the  given  log.,  4.031718 

Subtract  log.  of  10750  (Table  II.),  which  is  4.031408 

.000310 

This  difference  is  less  than  any  other  logarithm  in  Table  II. 
Therefore,  we  can  obtain  no  other  factor,  but  we  can  correct 
10750. 

The  next  greater  logarithm  in  the  table  is  that  of  10760, 
which  is  .031812 ;  the  tablular  diff.  is  .000404. 

Therefore,  .000404  :  000310  :  :  10  :  correction. 

Or,  404  :  310  :  :   10  :  correction  =  7.673. 


LOGARITHMS.  255 

3.  The  logarithm  of  the  British  gallon  is  2.442909;  a  cubic 
inch  being  the  unit.  How  many  cubic  inches  does  it  contain  ? 

Ans.  277.27+ 

(Art.  151^.)  Logarithms  are  exponents;  and  as  powers  and 
roots  may  be  expressed  by  exponents,  therefore,  logarithms 
may  be  used  for  finding  powers  and  roots. 

To  extract  the  square  root  of  any  number,  a,  we  divide  its 

exponent  (one  understood),  by  2,  and  the  result  is  a? .     To 

extract  the  third  root,  fourth  root,  &c.,  we  divide  the  exponent 
by  3,  4,  &c.,  as  the  case  may  be. 

The  exponent  of  a  number  is  the  logarithm  of  that  number. 
Hence,  to  extract  roots  by  logarithms,  we  have  the  following 
rule : 

RULE.  Take  the  logarithm  of  the  number  and  divide  it  by  2  for 
the  square  root,  by  3  for  the  third  root,  by  4  for  the  fourth  root, 
and  so  on. 

The  quotient  will  be  the  logarithm  of  the  root  sought,  and  the 
number  corresponding  to  this  logarithm  will  be  the  root  itself.  * 

(Art.  15U.)  If  x=a3  Then  log.  x—2>  log.  a. 

By  hypothesis,  x=a.a.a. 

By  taking  the  logarithm  of  each  member,  we  have 

log.  x=  log.  a  +  log.  a  -j-  log.  a  =  3  log.  a.     ( 1 ) 

Therefore,  in  general  terms, 

If  #=an  log.  x=n  log.  ff.(2) 

Now,  let  n  be  a  fraction,  then       x—cfo. 
Cube  both  members  ;  then  x3=a.     That  is,  x.x.x=a. 
By  the  property  of  logarithms  we  have 

log.  x  +  log.  x  +  log.  x  =  log.  a. 
Or,  3  log.  x  =  log.  a. 

Or,  log.  x  =  i  log.  a.  (3) 

All  these  equations  conform  to  the  rule  just  given. 

*  Examples  can  be  taken  out  of  any  Arithmetic. 


256  ELEMENTS  OF  ALGEBRA. 

USE    AND    APPLICATION    OF    LOGARITHMS. 

(Art.  152.)  The  sciences  of  trigonometry,  mensuration,  and 
astronomy  alone,  can  develop  the  entire  practical  utility  of  lo- 
garithms. The  science  of  algebra  can  only  point  out  their 
nature,  and  the  first  principles  on  which  they  are  founded.  To 
explain  their  utility,  we  must  suppose  a  table  of  logarithms  formed, 
corresponding  to  all  possible  numbers,  and  by  them  we  may  re- 
solve such  equations  as  the  following  : 

1.  Given  2*=  10  to  find  the  value  of  x. 

If  the  two  members  of  the  equation  are  equal,  the  logarithms 
of  the  two  members  will  be  equal,  therefore  take  the  logarithm 
of  each  member  ;  but  as  a?  is  a  logarithm  already,  we  shall  have 
v  log.  2=log.  10. 


2.  Given    (729)*  =3,  to  find  the  value  of  x. 

Raise  both  members  to  the  x  power,  and  3X=729=93, 
Or  3Z=36.     Hence,  #=6. 

3.  Given  «r+&*=c,   and  ax  —  bv=d,  to  find  the  values  of  x 
and  y. 

By  addition,  2ax=c-\-d.     Put  c-\-d=2m  ; 

Then       az=m.      Take  the  logarithm  of  each  mem- 

loff.  m 
her,    and    x  log.  «=log.  m,  or  x=^  -  . 

By   subtracting  the  second  equation    from   the   first   and 
making  c  —  rf=2n,  we  shall  find  2/==f"^"~l* 

3 

4.  Given    (2  1  6)  x  =18,  to  find  the  value  of  x. 

9  log.  6 

Ans.     x=-r-^~- 
log.  12 

9.  Given  —  ^=e,  to  find  the  value  of  x. 


losf.  m — log.  a  /fix 

Jtns.  x=—2— p.-    m  being  equal  to  (rfe-f  c] 


LOGARITHMS.  257 

£ 

6.  Given  43  =16,  to  find  the  value  of  x.  rftis.     x=Q. 

7»  Given  6*=  — - — —  to  find  the  value  of  x. 

18  log.24-{-log.l7--3  log.71 

Ans.     x— — ^ — . 

3  log.  6 

8.  Required  the  result  of  23.46  multiplied  by  7.218,  and  the 
product  divided  by  11.17. 

OPERATION. 

23.46 log.  1.37033 

7.218 log.  0.85842 

Sum 2.22875 

11.17          Subtr log.  1.04805 


Result,.  .  .  .     15.16  •  •  •  .log.  1.18070. 

N".  B.  The  log.  of  a  vulgar  fraction  is  found  by  subtracting 
the  log.  of  its  denominator  from  the  log.  of  its  numerator. 

For  instance,  T°T  is  simply  9  divided  by  11,  and  division  in 
logarithms  is  performed  by  subtraction. 

Thus,  from  the  log.  of  9,  ....    0.954343 

Subtract  log.  of  11,    ..    .    1.041393 

Log.  of  T°T  therefore  is  ....    — 1 .912850 

The  decimal  part  of  a  logarithm  is  never  minus,  but  the 
index  is  always  minus  when  the  number  is  less  than  unity. 
Hence,  the  logarithm  of  a  very  small  fraction  has  a  large  nega- 
tive index,  and  the  logarithm  of  0  is  minus  infinity.  The  loga- 
rithm of  10  is  1,  of  1  is  0,  of  .1  is  —I,  of  .01  it  is  — 2,  of  .001 
it  is  — 3,  and  so  on.  As  the  decimal  number  goes  down  to 
zero,  the  index  of  the  logarithm  goes  up  to  minus  infinity. 

The  previous  example  was  nothing  more  than  obtaining  the 
logarithm  of  a  common  fraction,  where  the  numerator  con- 
sisted of  two  factors.  That  example  might  have  been  stated 
thus : 

Find  the  log.  of  the  fraction,  (23-46)  (7-2*8) 

11.17 

22  Ans.   1.18070. 


258  ELEMENTS  OF  ALGEBRA. 

CHAPTER  V. 

COMPOUND  INTEREST. 

(Art.  153.)  Logarithms  are  of  great  utility  in  resolving  some 
questions  in  relation  to  compound  interest  and  annuities  ;  but  for 
a  lull  understanding  of  the  subject,  the  pupil  must  pass  through 
the  following  investigation  : 

Let  p  represent  any  principal,  and  r  the  interest  of  a  unit  of 
this  principal  for  one  year.  Then  1  -\-r  would  be  the  amouni 
of  $1,  or  J61.  Put^=l-fr. 

Now  as  two  dollars  will  amount  to  twice  as  much  as  one  dol 
lar,  three  dollars  to  three  times  as  much  as  one  dollar,  &c. 
Therefore,     1  :  Ji  :  :  A  :  JP=the  amount  in  2  years, 
And         1  :  Ji  :  :  JP  :  .#3=the  amount  in  3  years, 

<fec.   &c. 

Therefore,  «/?"  is  the  amount  of  one  dollar  or  one  unit  of  the 
principal  in  n  years,  and  p  times  this  sum  will  be  the  amount  for 
p  dollars.  Let  a  represent  this  amount  ;  then  we  have  this  gen- 
eral equation, 

pAn=a. 

In  questions  where  n,  the  number  of  years,  is  an  unknown 
term,  or  very  large,  the  aid  of  logarithms  is  very  essential  to  a 
quick  and  easy  solution. 

For  example,  what  time  is  required  for  any  sum  of  money 
to  double  itself,  at  three  per  cent,  compound  interest? 

Here  a=2/?,and  .#=(1.03),  and  the  general  equation  becomes 

/>(1.03)n=2/> 
Or  (1.03)"=2.       Taking  the  logarithms 


nk*.  (1.03)=log.  2,    or  n= 
years  nearly. 

2.  A  bottle  of  wine  that  originally  cost  20  cents  was  put  away 
for  two  hundred  years:  what  would  it  be  worth  at  the  end  of 
that  time,  allowing  5  percent,  compound  interest? 


ANNUITIES.  259 

This  question  makes  the  general  equation  staml  thus : 

(20  cts.  being  l  of  a  dollar)  i(1.05)200=  a 
Therefore  (1.05)200=5« 

Taking  the  logarithms        200  log.  (1.05)=log.5-{-log.  a 
Hence  log.  0=200  log.  (1.05)— log.  5.     JJns.     $3458.10 

3.  A  capital  of  $5000  stands  at  4  per  cent,  compound  interest; 
what  will  it  amount  to  in  40  years  ?  rfns.     $24005.10. 

4.  In  what  time  will   $5  amount  to  $9,  at  5  per  cent,  com- 
pound interest  ?  Jlns.     12.04  years. 

5.  A   capital  of  $1000   in   6   years,  at  compound  interest, 
amounted  to  $1800;  what  was  the  rate  per  cent? 

Ans.     log.  (l+r)=log*  L8      or      10T\  nearly 

6.  A  certain  sum  of  money  at  compound  interest,  at  4  per 
cent,  for  four  years,  amounted  to  $350. 95| ;  what  was  the  sum? 

fins.     $300. 

7.  How  long  must  $3600  remain,  at  5  per  cent,  compound  in- 
terest to  amount  to  as  much  as  $5000,  at  4  per  cent,  for  12  years  1 

Am.    16  years,  nearly. 

ANNUITIES. 

(Art.  154.)   An   annuity  is   a  sum   of  money   payable   peri 
odically,  for  some  specified  time,  or  during  the  life  of  the  re 
ceiver.     If  the  payments  are  not  made,  the  annuity  is  said  to  h 
in  arrear,  and  the  receiver  is  entitled  to  interest  on  the  several 
payments  in  arr3ar. 

The  worth  of  an  annuity  in  arrear,  is  the  sum  of  the  several 
payments,  together  with  compound  interest  on  every  payment 
after  it  became  due. 

On  this  definition  we  proceed  to  investigate  a  formula  to  be 
applied  to  calculations  respecting  annuities. 

Let   p   represent   the    annual    principal    or   annuity   to    be 


2GO  ELEMENTS  OF  ALGEBRA. 

paid,  and  \-\-r=*fl,  the  amount  of  annuity  of  principal  for  one 
year,  at  the  given  rate  r. 

Let  n  represent  the  number  of  years,  and  put  A'  to  represent 
the  entire  amount  of  the  annuity  in  arrear. 

It  is  evident,  that  on  the  last  payment  due,  no  interest  could 
accrue,  and  therefore  the  sum  will  be  p.  The  preceding  pay- 
ment will  have  one  year's  interest  ;  it  will  therefore  be  pJl  ;  the 
payment  preceding  that  will  have  two  years'  compound  interest  ; 
and,  of  course,  will  be  represented  by  pJP.  (Art.  153.)  Hence 
the  whole  amount  of  A'  will  be 


,  &c.,  to 
This  is  a  geometrical  series,  and  its  sum  (Art.  120.)  is 


This  general  equation  contains  four  quantities,  .#',  p,  r,  and  n  , 
any  three  of  them  being  given  in  any  question,  the  others  can  be 
found,  except  r. 

EXAMPLES. 

1.  An  annuity  of  $50  has  remained  unpaid  for  6   years,  at 
compound  interest  on  the  sums  due,  at  6  per  cent.,  what  sum  is 
now  due  ? 

By  the  general  equation, 

50[(t.06)'-l] 

~:o6~ 

Faking  the  log.  of  both  members,  we  have 

log.  A'=  log.  50-flog.  [(1.06)6—  1]—  log.  .06. 

The  value  of  (1.Q6)6,  as  found  by  logarithms,  is  1.41852,  from 
which  subtract  1,  as  indicated,  and  take  the  log.  of  the  decimal 
number  .41852,  we  then  have 

log.  ^'=1.  69897  +(—  1.62172)—  (—  2.778151)=2.54218, 
From  which  we  find,  .#'=$348.56     Jim. 

2.  In  what  time  will  an  annuity  of  $20  amount  to  $1000,  at 
4  per  cent.,  compound  interest  ? 


ANNUITIES.  261 

The  equation  applied,  we  have 


Dividing  by  20,  and  multiplying  by  .04,  we  have 
2=(1.04)»—  1     or     (1.04)»=3. 

log.  3        .477121 


3.  What  will  an  annuity  of  $50  amount  to,  if  suffered  to 
remain  unpaid  for  twenty  years,  at  3|  per  cent,  compound  in- 
terest? Ans.     $1413.98. 

4.  What  is  the  present  value  of  an  annuity  or  rental  of  $50 
a  year,  to  continue  20  years,  discounting  at  the  rate  of  3£  per 
cent.,  compound  interest  ? 

N.  B.  By  question  3d,  we  find  that  if  the  annuity  be  not  paid 
until  the  end  of  20  years,  the  amount  then  due  would  be  $1413.98. 
If  paid  now,  such  a  sum  must  be  paid  as,  put  out  at  compound 
interest  for  the  given  rate  and  time,  will  amount  to  $1413.98. 

Now  if  we  had  the  amount  of  $1  at  compound  interest  for  20 
years,  at  3j  per  cent.,  that  sum  would  be  to  $1  as  $1413.98  is 
to  the  required  sum,  $710.62. 

(Art.  1  55.)  To  be  more  general,  let  us  represent  the  present 
worth  of  an  annuity  by  P.  By  (Art.  153.)  the  amount  of  one 
dollar  for  any  given  rate  and  time,  is  Jin  ;  A  being  \-\-r  and  n 
the  number  of  years.  By  (Art.  154.)  the  value  of  any  annuity 
p  remaining  unpaid  for  any  given  time,  n  years,  at  any  rate  of 

pAn  —  p 
compound  interest  r,  is  --  —  or  Jl'. 

Now  by  the  preceding  explanation  we  may  have  this  propor- 
tion : 

Jl*  :l::J':P,     or    P=~-  ......  (1) 

Hence,  to  find  the  present  worth  of  an  annuity,  we  have  this 
RULE.     Divide  the  amount  of  the  annuity  supposed  unpaid 

for  the  given  number  of  years,  by  the  amount  of  one  dollar  for 

the  same  number  of  years. 


2  ELEMKXTS  OF  ALCEBRA. 

If  in   equation  (1)  we  put  the  value  of  .#,  we   shall  have 

^n—E Lm     Divide  both  members  by  .#*,  and  we  have 


(2) 


r 

This  last  equation  will  apply  to  the  following  problems  : 

5.  The  annual  rent  of  a  freehold  estate  is  p  pounds  or  dollars, 
to  continue  forever.     What  is  the  present  value  of  the  estate, 
money  being  worth  5  per  cent.,  compound  interest  ? 

Here,  as  n  is  infinite,  the  term,  •—  becomes  0,  and  equation 

(2)  becomes  P=L—JL^=2Qp  •  that  is,  the  present  value  of  the 
estate  is  worth  20  years'  rent. 

6.  The  rent  of  an  estate  is  $3000  a  year ;  what  sum  could 
purchase  such  an  estate,  money  being  worth  3  per  cent.,  com- 
pound interest  ?  Jim.     $100000. 

7.  What  is  the  present  value  of  an  annuity  of  $350,  assigned 
for  8  years,  at  4  per  cent.  ?  Jlns.     $2356.46. 

8.  A  debt  due  at  this  time,  amounting  to  $1200,  is  to  be  dis- 
charged in  seven   annual   and   equal  payments  ;    what   is   the 
amount  of  these  payments,  if  interest  be  computed  at  4  per  cent.  ? 

Ans.     $200,  nearly. 

9.  The  rent  of  a  farm  is  $250  per  year,  with  a  perpetual  lease. 
How  much  ready  money  will  purchase  said  farm,  money  being 
worth  7  per  cent,  per  annum  1  rfns.     $3571 7 

10.  An  annuity  of  $50  was   suffered  to  remain  unpaid  for 
20  years,  and  then  amounted  to  $1413.98;  what  was  the  rate 
per  cent.,  at  compound  interest  ? 

N.  B.  This  question  is  the  converse  of  problem  3,  and,  ol 
course,  the  answer  must  be  3i  per  cent.  But  the  general  equa- 
tion gives  us 


GENERAL  THEORY  OF  EQUATIONS.  %63 

Or  28.2796  JH^2— ; 

an  equation  from  which  it  is  practically  impossible  to  obtain  r, 
except  by  successive  approximations. 


SECTION   VII. 

CHAPTER  I. 
GENERAL  THEORY  OF  EQUATIONS. 

(Art.  156.)  In  (Art.  101.)  we  have  shown  that  a  quadratic 
equation,  or  an  equation  of  the  second  degree,  may  be  conceived 
to  have  arisen  from  the  product  of  two  equations  of  the  first  de- 
gree. Thus,  if  x=a,  in  one  equation,  and  x=b  in  another 
equation,  we  then  have 

x — a=0, 
and       x — &=0; 


By  multiplication,       x2  —  (a-{-b}x-{-ab=Q. 

This  product  presents  a  quadratic  equation,  and  its  two  roots 
are  a  and  b. 

If  one   of  the  roots  be   negative,  as  x=  —  «,  and  x=b,  the 
resulting  quadratic  is 

xz-\-(a—b)x—ab=0. 

If  both  roots  be  negative,  then  we  shall  have 


Now  let  the  pupil  observe  that  the  exponent  of  the  highest 
power  of  the  unknown  quantity  is  2  ;  and  there  are  two  roofs. 
The,  coefficient  of  the  first  power  of  the  unknown  quantity  is 
the  algebraic  sum  of  the  two  roots,  with  their  signs  changed  ; 
and  the  absolute,  term,  independent  of  the  unknown  quantity, 
is  the  product  of  the  roots  (the  sign  conforming  to  the  rules  of 
multiplication}. 


264  ELEMENTS  OF  ALGEBRA. 

When  the  coefficients  and  absolute  term  of  a  quadratic  are  not 
largo,  and  not  fractional,  we  may  determine  its  roots  by  inspec- 
tion, by  a  careful  application  of  these  principles 

EXAMPLES. 

Given  #*— 20#-f96=0,  to  find  x. 

The  roots  must  be  12  and  8,  for  no  other  numbers  will  make 
— 20,  signs  changed,  and  product  96. 

Given  yz — 6y — 55=0   to  find  y.  Roots  11  and — 5. 

Given  a? — 6#— 40=0    to  find  x.  Roots  10  and  — 4. 

Given  s?+§x — 91=0   to  find  x.  Roots  7  and  — 13. 

Given  y* — 5y — 6=0   to  find  y.  Roots  6  and  — 1. 

Given  yz-\-\'2y— 589=0   to  find  y. 

Here  it  is  not  to  be  supposed  that  we  can  decide  the  values  of 
the  roots  by  inspection;   the  absolute  term  is  too  large;   but, 
nevertheless,  the  equation  has  two  roots. 
Let  the  roots  be  represented  by  P  and  Q. 
From  the  preceding  investigation 

P+$=-  12 (1) 

And  PQ=— 589 (2) 

By  squaring  eq.  (1)      P2-}-2PQ-{-Q2=       144 
4  times  eq. .  .  .  (2)  4PQ         =—2356 

By  subtraction,  p2—2PQ+Q2=     2500 

By  evolution,          P — #=±50 
But  P-f-£=— .12 

Hence  P=19  or  — 31,  and  Q=—3l  or  +19,  the  true 
roots  of  the  primitive  equation ;  and  thus  we  have  anothci 
method  of  resolving  quadratics. 

(Art.  157.)  In  the  same  manner  we  can  show  that  the  product 
of  three  simple  equations  produce  a  cubic  equation,  or  an  equa- 
tion of  the  third  degree.  Conversely,  then,  an  equation  of  the 
third  degree  has  three  roots. 

The  three  simple  equations,  #=«,   x=b,   x=c,*  may  be  put 

*  Of  course,  x  cannot  equal  different  quantities  at  one  and  the  same  time 
and  these  equations  must  not  be  thus  understood. 


GENERAL  THEORY  OF  EQUATIONS.       265 

in  the  form  of  x  —  a=Q,   x  —  &=0,   and   x  —  c=0,  and  the  pro- 
duct of  these  three  give 

(x  —  a}(x  —  b]  (x  —  c)=0  ; 
and  by  actual  multiplication,  we  have 
x3—  a 


If  one  of  the  roots  be  negative,  as  x=  —  c,  or  a?-j-c=0,  the 
product  or  resulting  cubic  will  be 


If  two   of  them   be  negative,   as  x=  —  b  and  #==  —  e,  the 
resulting  cubic  will  be 

a^+C^+c  —  a)x2-f(6c  —  ab  —  ac}x  —  abc=0. 
If  all  the  roots  be  negative,  the  resulting  cubic  will  be 


Every  cubic  equation  may  be  reduced  to  this  form,  and  con- 
ceived to  be  formed  by  such  a  combination  of  the  unknown  term 
and  its  roots. 

By  inspecting  the  above  equations,  we  may  observe 

1st.  The  first  term  is  the  third  power  of  the  unknown 
quantity. 

2d.  The  second  term  is  the  second  power  of  the  unknown 
quantity,  with  a  coefficient  equal  to  the  algebraic  sum  of  the 
roots,  with  the  contrary  sign. 

3d.  The  third  term  is  the  first  power  of  the  unknown 
quantity,  with  a  coefficient  equal  to  the  sum  of  all  the  products 
which  can  be  made,  by  taking  the  roots  two  by  two. 

4th.  The  fourth  term  is  the  continued  product  of  all  the  roots, 
with  the  contrary  sign. 

It  is  easy,  then,  to  form  a  cubic  equation  which  shall  have 
any  three  given  numbers  for  its  roots. 

Assuming  x  for  the  unknown  quantity,  find  an  equation  which 
shall  have  1,  2  and  3  for  its  roots. 

tins.     a*—(l+2+3)3M-(2+  3+6)*—  6=0  ; 
Or 


Find  the  equation  which  shall  have  2,  3,  and  —  4  for  its  roots. 
23  Ans.    x*—  9?—  14a?+24=0. 


266  ELEMENTS  OF  ALGEBKA. 

Find  the  equation  which  shall  have  —  3,  —  4y  and  +7  for  its 
roots.  Am.  x3±0a;2—  37z—  84=0, 

Or   x*—  37#—  84=0. 

These  four  general  cases  of  cubic  equations  may  all  be  repre- 
sented by  the  general  form. 

Thus:  xs-\-pxz+qx+r=0,  .........   (1) 

(Art.  158.)  When  the  algebraic  sum  of  three  roots  is  equal  U 
zero,  equation  (1)  takes  the  form  of 

x*+qx+r=0  .............   (2) 

Equation  (1)  is  a  regular  cubic,  and  is  not  susceptible  of  a 
direct  solution,  by  Cardan's  rule,  until  it  is  transformed  into 
another  wanting  the  second  term,  thus  making  it  take  the  form 
of  equation  (2).  To  make  this  transformation,  conceive  one  of 
the  roots,  or  x,  in  equation  (1),  represented  by  u-{-v* 

Then  x*=u*+3uzv+3uvz  -f-u3 

px*=       pu2  -\-2puv-\-pv2 
qx  =  qu   -\-qv 

r  =  r 

By  addition,  and  uniting  the  second  member  according  to  tho 
powers  of  u,  we  shall  have 


for  the  transformed  equation.     But  the  object  was  to  make  such 
a  transformation  that  the  resulting  equation  should  be  deprived 
of  its  second  power  ;   and  to  effect  this,  it  is  obvious  that  we 
must  make  the  coefficient  of  u2  equal  zero,  or  3v-}-p=Q. 
Therefore,  v=  —  ip. 

Hence,  we  perceive  that  if  x,  in  the  general  equation  (1),  be 

put  equal  to  u—  ~,  there  will  result  an  equation  in  the  form  of 
o 

u*-\-qu-\-r=0,  or  the  form  of  equation  (2). 

As  x=u  —  ^,  and  if  «,  6,  and  c  represent  the  roots  of  equa- 
o 

tion  (1),  or  the  values  of  x,  the  roots  of  (2),  or  values  of  u  will  be 

and 


GENERAL  THEORY  OF  EQUATIONS.       267 

EXAMPLES. 

1.  Transform  the  equation  ar3 — 9x?-{-26x — 30=0,  into  another 
wanting  the  second  term. 

By  the  preceding  investigation,  we  must  assume 

x—u-}-3.     Here  p=— 9  ;  therefore,  — sp=3. 


=  26w-f78 

—30  =  —30 

Sum,  0  =w3  — u —  6=0,    the  equation  required. 

2.  Transform  the  equation  a?3— 6rJ+10a>— 8=0,  into  another 
not  containing  the  square  of  the  unknown  quantity. 

Put  x=u+2.  Result,    u3 — 2w — 4=0. 

3.  Transform   xs — 3xz-}-6x — 12=0,   into  another    equation, 
wanting  the  second  power  of  the  unknown  quantity. 

Put  a?=w+l.  Result,    w3+3w — 8=0. 

(Art.  159.  We  have  shown,  in  the  last  article,  that  any  icgular 
cubic  equation  containing  all  the  powers  of  the  unknown  quan- 
tity can  be  transformed  into  another  equation  deficient  of  the 
second  power  ;  and  hence  all  cubic  equations  can  be  reduced  to 
the  form  of 


We  represent  the  coefficient  of  x  by  3p,  and  the  absolute  term 
by  2</,  in  place  of  single  letters,  to  avoid  fractions,  in  the  course 
of  the  following  investigation. 

Now,  if  we  can  find  a  direct  solution  to  this  general  equation, 
it  will  be  a  solution  of  cubic  equations  generally. 

The"  value  of  x  must  be  some  quantity  ;  and  let  that  quantity, 
whatever  it  is,  be  represented  by  two  parts,  v-\-y,  or  let 
x=v-\-y.  Then  the  equation  becomes 


By  expanding  and  reducing,  we  have 


Now  as  we  have  made  an  arbitrary  division  of  x  into  two  parts, 
v  and  y   we  can  so  divide  it,  that 


268  ELEMENTS  OF  ALGEBRA. 

This  hypothesis  gives 

v3+y*=  2q, (.#) 

And  Vy=—p, (J9) 

Here  we  have  two  equations,  (A)  and  (/?),  containing  two  un- 
known quantities,  similarly  involved,  which  admit  of  a  solution 
by  quadratics.  (Art.  108.)  Hence  we  obtain  v  and  y,  and  their 
algebraic  sum  is  x. 

From  equation  (B], 


This  substituted  in  equation  («$),  gives 

Or,  VQ — 2yu3=p3,  a  quadratic. 

Hence  v- 

And  y. 

Or,  as  y- 


Therefore     x=(q+  jf+fi}*  +(q— 


Or, 


These  formulas  are  familiarly  known,  among  mathematicians,  aa 
Cardan's  rule. 

(Art.  160.)  When  p  is  negative,  in  the  general  equation,  and 
its  cube  greater  than  <?2,  the  expression  Jq*  —  ps  becomes  imagi-' 
nary  ;  but  we  must  not  conclude  that  the  value  of  #  is  therefore 
imaginary  ;  for,  admitting  the  expression  ,J(f  —  -jo3  imaginary,  it 

can  be  represented  by  aj  —  1,  and  the  value  of  a?,  in  equation 
(C),  will  be 


(JENERAL  THEORY  OF  EQUATIONS.  269 


Now  by  actually  expanding  the  roots  of  these  binomials  by  the 
binomial  theorem,  and  adding  their  results  together,  the  terms 
containing  J  —  1  will  destroy  each  other,  and  their  sum  will  be 
a  real  quantity  ;  and,  of  course,  the  value  of  x  will  become  real. 
If  in  any  particular  case  it  becomes  necessary  to  make  the  series 

converge,  change  the  terms  of  the  binomial,  and  make  —J  —  1 
stand  first,  and  1  second. 

EXAMPLES. 
Given  x9  —  6a?=5.6,  to  find  the  value  of  x. 

Here,  3p=  —  6,  and  2?—  5.6,  or  p=  —  2,  and  #=2.8. 


Then 


*=(2.8+/7.84— 8)  +(2.8—77.84—8)  ,  by  equation  (C) 


Or    tf=(2.8+.4,/—  1)  +(2.8—  .47—  1) 

Or  " 


Expand  the  binomials  by  the  binomial  theorem,  (Art.  135.), 
i  d  for  the  sake  of  brevity,  represent  4  J  —  1  by  b; 

Then     &*=—     ,    and    fl«='x- 


270  ELEMENTS  OF  ALGEBRA 


=24-0.004535—0.000034=2.0045. 
Therefore, 


3-—=2.0045  or  a?=(2.0045)(2.8)=2.8256,  nearly 
(Art.  161.)  Every  cubic  equation  of  the  form  of 


has  three  roots,  and  their  algebraic  sum  is  0,  because  the  equa- 
tion is  wanting  its  second  term.     (Art.  157.) 

If  the  roots  be  represented  by  «,  b,  and  c,  we  shall  have 


If  any  two  of  these  roots  are  equal,  as  b=c,  then  a=  —  2b  (1), 
and  ab2=±q  (2).  Putting  the  value  of  a  taken  from  equation 
(1),  into  equation  (2),  and  we  have  —  263=±gr. 

Hence,  in  case  of  there  being  two  equal  roots,  such  roots  must 
each  equal  the.  cube  root  of  one  half  the  quantity  represented 
byq. 

EXAMPLES. 

The  equation  re3  —  48^=128  has  two  equal  roots;  what  are 
.he  roots  ? 

Here,  —  263,=128,  or  bs=—  64;  therefore,  b=  —  4. 

Two  of  the  roots  are  each  equal  to  —  4,  and  as  the  sum  of  the 
three  roots  must  be  0,  therefore  —  4,  —  4,  -\-8,  must  be  the 
three  roots. 

If  the  equation  a:3  —  27#=54  have  two  equal  roots,  what  are 
the  roots  ?  Am.  —  3,  —  3,  and  +6. 

Either  of  these  roots  can  be  taken  to  verify  the  equation  ;  and 
if  they  do  not  verify  it,  the  equation  has  not  equal  roots. 

(Art.  162.)  If  a  cubic  equation  in  the  form  of 


GENERAL  THEORY  OF  EQUATIONS.       271 

have   two   equal   roots,  each   one   of   the   equal   roots  will  bo 
equal  to 


The  other  root  will  be  twice  this  quantity  subtracted  from  p, 
because  the  sum  of  the  three  roots  equal  p.  (Art.  157.) 

This  expression  is  obtained  from  the  consideration  that  the 
tfirce  roots  represented  by  «,  6,  and  c,  must  form  the  folloAving 
equations:  (Art.  157.) 

a+b+c=p  .........  (1) 

ao-{-ac-\-bc=q  .........  (2) 

abc—r  .........  (3) 

On  the  assumption  that  two  of  these  roots  are  equal,  tha*  is, 
a=b,  equations  (1)  and  (2)  become 

2a-\-c=p  ........  (4) 

And  a2-{-2ac=q  ........  (5) 

Multiply  equation  (4)  by  2«,  and  we  have 

4a2-\-2ac=2ap  .......  (7) 

Subtract  (5)     «2+2oc=     q  .......  (8) 

And  we  have   3a2  =2ap  —  q. 

This  equation  is  a  quadratic,  in  relation  to  the  root  or,  and  a 
solution  gives  a=i(p±.Jp2  —  3q). 

(Art.  163.)  A  cubic  equation  in  the  form  of  xs±px=±q  can 
be  resolved  as  a  quadratic,  in  all  cases  in  which  q  can  be  resolved 
into  two  factors,  m  and  rc,  of  such  a  magnitude  that  m2-\-p=n. 

For  the  values  of  p  and  </,  in  the  general  equation,  put  the 
assumed  values,  mn=q,  and  p=n  —  mz. 

Then  we  have         x3-{-nx  —  m2x=mn. 

Transpose  —  mzx,  and  then  multiply  both  members  by  x,  and 
=  mzxz-{-mnx. 


Add  --  to  both  members,  and  extract  square  root  ; 

Then   x*-\~-—mx-\--.     Drop  -,  and  divide  by  a?,  and  x-=m. 


272  ELEMENTS  OF  ALGEBRA. 

Therefore,  if  such   factors  of  q  can  be  found,  the  equation  is 
already  resolved,  as  x  will  be  equal  to  the  factor  m. 

EXAMPLES. 

1.  Given  x?+6x=S&,  to  find  the  values  of  x 

Here,  wm=88=4X22,     42-j-6=22.     Hence,  x=4 

2.  Given  x3-\-3x=l4,  to  find  one  value  of  x. 

2X7=14,     22+3=7.     Hence,  x— 2. 

U.  Given  #3-|-6#=45,  to  find  one  value  of  x.       *fl.ns.  x=3. 

4.  Given  x3 — 13;r= — 12,  to  find  x.  rfns.     x=3. 

5.  Given  7/3+48?/=104,  to  find  y.  Jlns.     y=2. 

In  the  above  examples  we  have  given  only  one  answer,  or  one 
root ;  but  we  have  more  than  once  observed,  that  every  equation 
of  the  third  degree  must  have  three  roots.  Take,  for  example, 
the  4th  equation.  We  have  found,  as  above,  one  of  its  roots  to 
be  3.  Now  we  may  conceive  the  equation  to  have  been  the 
product  of  three  factors,  one  of  which  was  (x — 3) ;  therefore  the 
equation  must  be  divisible  by  x — 3  without  a  remainder,  (other- 
wise 3  cannot  be  a  root)  ;  and  if  we  divide  the  equation  by  x — 3, 
the  quotient  must  be  the  product  of  the  other  two  factors. 

Thus,       a:— 3)*3—  I3x+12(xz-\-3x— 4 


2—  I3x 


—  4H-12 

By  putting  x*-\-3x  —  4=0,  and  resolving  the  equation,  we  find 
a?=l,  or  —  4,  and  the  three  roots  are  1,  3,  —  4.     Their  sum  is 
0,  as  it  should  be,  as  the  equation  is  deficient  of  its  second  term. 
In  the  general  equation 


If  p  and  q  are  each  equal  to  0,  at  the  same  time,  the  equation 
becomes  ai3-j-r=0,  a  binomial  equation. 


GENERAL  THEORY  OF  EQUATIONS.       373 

Every  binomial  equation  has  as  many  roots  as  there  are 
units  in  the  exponent  of  the  unknown  quantity. 

Thus  a;3+8=0,  and  a3— 8=0,  or  o;3+l=0,  and  a;3— 1=0, 
&c.,  are  equations  which  apparently  have  but  one  root,  but  a 
full  solution  will  develop  three. 

Take,  for  example,  a? =8 

By  evolution,  x  =2 


4x—S 
4x—S 


Now  by  putting  a;2+2^-}-4=0,  and  resolving  the   equation, 

we  find  x= — 1  +  V — **»  an(^  x~ — * — >/"~~^  »  anc^  tne  tnree 
roots  of  the  equation  Xs — 8  =  0,  are  2,  — 1  +  ^/ — 3,  and 
— 1 — J — 3,  two  of  them  imaginary,  but  either  one,  cubed,  will 
give  8. 

The  three  roots  of  the  equation  ic3+l=0,  are 

1  ,  1    . 1      1 


CHAPTER  II. 
GENERAL  THEORY  OF  EQUATIONS— CONTINUED. 

(Art.  164.)  In  the  last  Chapter  we  confined  our  investigations 
to  equations  of  the  second  and  third  degrees ;  and  if  they  are 
well  understood  by  the  pupil,  there  will  be  little  difficulty,  in 
future,  as  many  of  the  general  properties  belong  to  equations  of 
every  degree. 

All  the  higher  equations  may  be  conceived  to  have  been  formed 
by  the  multiplication  of  the  unknown  quantity  joined  to  each 
of  the  roots  of  the  equation  with  a  contrary  sign,  as  shown  in 
(Art.  157.). 


274  ELEMENTS  OF  ALGEBRA. 

Let  a,  b,  c,  d,  e,  &c.,  be  roots  of  an  equation,  and  a; 
its  unknown  quantity,  then  the  equation  may  be  formed  by  the 
product  of  (x  —  a}(x  —  b)(x  —  c),  &c.,  which  product  we  may 
represent  by 


Now  it  being  admitted  that  equations  ran  be  thus  formed  by 
the  multiplication  of  the  unknown  quantity  joined  to  its  roots, 
conversely,  when  any  of  its  roots  can  be  found,  such  root,  with 
its  contrary  sign  joined  to  the  unknown  term,  will  form  a  com- 
plete divisor  for  the  equation  ;  and  by  the  division  the  equation 
will  be  reduced  one  degree,  and  conversely. 

If  any  quantity,  connected  to  the  unknown  quantity  by  the 
sign  plus  or  minus,  divide  an  equation  without  a  remainder, 
such  a  quantity  may  be  regarded  as  one  of  the  roots  of  the 
equation. 

The  product  of  all  the  roots  form  the  absolute  term  U. 

(Art.  165.)  Every  equation  having  unity  for  the  coefficient 
of  the  first  term,  and  all  the  other  coefficients,  whole  numbers, 
can  have  only  whole  numbers  for  its  commensurable*  roots. 

This  being  one  of  the  most  important  principles  in  the  theory 
of  equations,  its  enunciation  should  be  most  clearly  and  distinctly 
understood.  Such  equations  may  have  other  roots  than  whole 
numbers  ;  but  its  roots  cannot  be  among  the  definite  and  irre- 
ducible fractions,  such  as  f  ,  J,  U  ,  &e.  Its  other  roots  must  be 

among  the  incommensurable  quantities,  such  as  J2,  (3)%  &c., 
i.  e.,  surds,  indeterminate  decimals,  or  imaginary  quantities. 

To  prove  the  proposition,  let  us  suppose  j-  a  commensurable 
but  irreducible  fraction,  to  be  a  root  of  the  equation 


A,  B,  «fec.,  being  whole  numbers. 
Substituting  this  supposed  value  of  x,  we  have 


*  Commensurable  numbers  are  all  those  that  measure  or  can  be  measured  by 
unity  ;  hence,  all  whole  numbers  and  definite  fractions  are  commensurable.  — 
Surds,  and  imaginary  quantities,  are  incommensurable. 


GENERAL  THEORY  OF  EQUATIONS.       275 

Transpose  all  the  terms  but  the  first,  and  multiply  by  bm~},  and 
we  have 

am 

y- 

Now,  as  a  and  b  are  prime  to  each  other,  b  cannot  divide  a, 
or  any  number  of  times  that  a  may  be  taken  as  a  factor,  for  j 

being  irreducible,  -  X  a  is   also  irreducible,  as  the  multiplier  a 

a? 
will  not  be  measured  by  the  divisor  b  ;  therefore  y    cannot    be 

expressed  in  whole  numbers.     Continuing  the  same  mode  of 

am 
reasoning,  y  cannot  express  whole  numbers,  but  every  term  in 

the  other  member  of  the  equation  expresses  whole  numbers. 
Hence,  this  supposition  that  the  irreducible  fraction  •=•  is  a  root 

of  the  equation,  leads  to  this  absurdity,  that  a  series  of  whole 
numbers  is  equal  to  another  quantity  that  must  contain  a  fraction. 

Therefore,  we  conclude  that  any  equation  corresponding  to 
these  conditions  cannot  have  a  definite  commensurable  fraction 
among  its  roots. 

(Art.  166.)  Any  equation  having  fractional  coefficients,  can  be 
transformed  to  another  in  which  the  coefficients  are  all  whole 
numbers,  and  that  of  the  first  term  unity. 
For  example,  take  the  equation 


m      n     p 

Assume   ar=-£-,  and  put  this  value  of  x  in  the  equation, 
mnp 


And       _ 


_.. 
mn*p    p 

Multiply  every  term  by  m*n*p*,  and  we  have 


276  ELEMENTS  OF  ALGEBRA. 

When  m,  n,  and  p  have  common  factors,  we  may  put  x  equal 
to  y  divided  by  the  least  common  multiple  of  these  quantities, 
as  in  the  following  examples: 

Transform    the    equation    y*-\  --  .+  ---  h-=0,  into  another 

pm     m    p 

which  shall  have  no  fractional  coefficients,  and  that  of  the  first 
term  unity. 

To  effect  this,  it  is  sufficient  to  put  x=-^—.     With  this  value 

pm 

of  x  the  equation  becomes 


__.=. 

pV    p*m3    pm2    p 
Multiplying  every  term  by  j93m3,  we  obtain 


for  the  transformed  equation  required. 
Transform  the  equation  x*-\  —  —  -  1  —  'T"i"nA~^"Tn==^9  into  an- 

other, having  no  fractional  coefficients. 

Result,    y+20?/3+18.24y4-7(24)2y-}-2(24)3=0. 

(Art.  167.)  Now  as  every  commensurable  root  consists  of 
whole  numbers,  and  as  the  coefficients  are  all  whole  numbers, 
each  term  of  itself  consists  of  whole  numbers,  and  the  commen- 
surable roots  are  all  found  among  the  whole  number  divisors  of 
the  last  term  ;  and  if  these  divisors  are  few  and  obvious,  those 
answering  to  the  roots  of  the  equation  may  be  found  by  trial.  If 
the  factors  are  numerous,  we  must  have  some  systematic  method 
of  examining  them,  such  as  is  pointed  out  by  the  following  rea- 
soning : 

Take  the  equation  *4+^3+j£arz+Ca?-j-Z)==(). 

Let  a  represent  one  of  its  commensurable  roots.  Transpose 
all  the  terms  but  the  last,  and  divide  every  term  by  a. 


But,  since  a  is  a  root  of  the  equation,  it  divides  D  without  a 


GENERAL  THEORY  OF  EQUATIONS.       277 

remainder,  the  left  hand  member  of  this  last  equation  is  therefore 
a  whole  number,  to  which  transpose  C,  also  a  whole  number, 

and  represent  —  \-C  by  N. 

Then  N=—a?—rfa?—£a. 

Divide  each  term  by  a,  and  transpose  J3,  and  we  have 

-+£=—  a2—  da. 
a 

The  right  hand  member  of  this  equation  is  an  entire  quantity, 
(not  fractional),  therefore  the  other  member  is  also  an  entire 
quantity  ;  let  it  be  represented  by  N1,  and  the  equation  again 
divided  by  a. 

Then  —  =—  a—  A. 

a 

Transpose  —  A  ;  reasoning  the  same  as  before,  we  can  repre- 
sent the  first  member  by  JV",  and  we  then  have 


N" 
Divide  by  a,  and  —  =  —  1.     This  must  be  the  final  result,  in 

case  a  is  a  root. 

From  these  operations  we  draw  the  following  rule  for  deciding 
what  divisors  of  the  last  term  are  roots  of  an  equation. 

RULE.  Divide  the  last  term  by  the  several  divisors,  (each 
designated  by  «,)  and  add  to  the  quotient  the  coefficient  of  the 
term  involving  x. 

Divide  this  sum  by  the  divisors  (a),  and  add  to  the  quotient 
the  coefficient  of  the  term  involving  #*. 

Divide  this  sum  by  the  divisors  (a),  and  add  to  the  quotient 
the  coefficient  of  the  term  involving  x?. 

And  thus  continue  until  the  first  coefficient,  .#,  is  transposed, 
and  the  sum  divided  by  a  ;  the  last  quotient  will  be  minus  one, 
if  a  is  in  fact  a  root 


278 


ELEMENTS  OF  ALGEBRA. 


EXAMPLES. 

1.  Required  the  commensurable  roots  (if  any)  of  the  equation 

=0. 


N 

£• 

2                                       1 

(M 

0 

7 

1                                       1 

±» 

CD 

<N 

_u                                                                                          "T3 

O 

1 

7                i 

1 

7 

CO 

i 
2                                    1 

0* 

o> 

7 

* 

7 

7     +7 

i 

0) 

1 

CO 

^5      ift 

i-H 

,|     ,7    ,_ 

1 

7 

c* 

1—4 

7  ' 

CO     CO                '-''-*§ 

1         +    1      a 

•1 

"c 

^4 

00     QO 

05     05                                        .22 

ij 

7 

tl        I 

bo 

c* 

CD 

<N     f-t 

^UT  7  | 

.s 

CO 

7 

0 

? 

3 

!i 

1 

o 

a    2 

* 

1 

05 

7 

1 

|l 

o 

O^    "~^ 

CO 

T 

777 

1         1 

1  i 

e*    <M 

r-l        CD 

^                       ^ 

C  | 

1 

1  7 

i            e 

CD 

"°     oT 

»           a    ^ 
il           1           Is 

it 

.       1           1           1          1    t 

1       ||        11       11       &    | 

_>:                            rJ                                                                                                                                   r^              ^ 

o    -5 

O        M 

•sl 

CO 


EQUAL  ROOTS.  279 

2.  Required    the     commensurable    roots     of    the    equation 
ar5*—  6r4-lla:—  6=0.  dns.     1,2,3. 

3.  Required     the    commensurable    roots    of    the    equation 
s«_6;e2—  16^+21=0.  Am.     3  and  1 

Here  the  student  might  hesitate,  as  one  regular  term  of  the 
equation  is  wanting,  or  rather  the  coefficient  of  x*  is  0  .  hence, 
the  equation  is  #4dbOa;3  —  6cr2—  16H-21=0. 
Go  through  the  form  of  adding  0. 

4.  Required    the    commensurable    roots    of    the     equation 
a>4—  6a?34-5^+2#—  10=0.  tins.     —  1,  +5. 

As  the  commensurable  roots  are  only  two,  there  must  be  two 
incommensurable  roots  ;  and  they  can  be  found  by  dividing  the 
given  equation  by  a?-j-l,  and  that  quotient  by  x  —  5,  and  resolv- 
ing the  last  quotient  as  a  quadratic. 

EQUAL  ROOTS. 

(Art.  168.)  In  any  equation,  as 


the  roots  may  be  represented  by  «,  &,  c,  d,  e,  and  either  one,  put 
in  the  place  of  x,  will  verify  the  equation. 

Now,  let  y  represent  the  difference  between  any  two  roots,  as 
a  —  b  ;  then  y=a  —  &,  and  by  transposition  b-\-y=a.  But  as  a 
will  verify  the  equation,  it  being  a  root,  its  equal,  (b-\-y),  sub- 
stituted for  x,  will  verify  it  also.  That  is, 


By  expanding  the  powers,  and  arranging  the  terms  according 
to  the  powers  of  y,  we  have 
+106V 


Cb2+2Cby 

Db+Vy 

E 


=0. 


We  might  have  been  more  general,  and  have  taken  xm-{-Axm—1,  &c.,  for 
the  equation  ;  but,  in  our  opinion,  we  shall  be  better  comprehended  by  taking 
an  equation  definite  in  degree :  the  reasoning  is  readily  understood  as  general 


280  ELEMENTS  OF  ALGEBRA. 

Now,  as  I  is  a  root  of  the  equation,  the  first  column  of  this 
transformation  is  identical  with  the  proposed  equation,  on  sub- 
stituting the  root  b  for  x.  Hence,  the  first  column  is  equal  to 
xero  ;  therefore,  let  it  be  suppressed,  and  the  remainder  divided 
byy. 

We  then  have 


=0. 


2Cb 
D 


On  the  supposition  that  the  two  roots  a  and  b  are  equal,  y 
becomes  nothing,  and  this  last  equation  becomes 


As  b  is  a  root  of  the  original  equation,  x  may  be  written  in 
place  of  b  ;  then  this  last  equation  is 

5zM-4^e3+3^+2Ca:+#=0.  .  .  .  .  .  (2) 

This  equation  can  be  derived  from  the  primitive  equation  by 
the  following 

RULE.  Multiply  each  coefficient  by  the  exponent  of  x,  and 
diminish  the  exponent  by  unity. 

Equation  (2)  being  derived  from  equation  (1),  by  the  above 
rule,  may  be  called  a  derived  polynomial. 

(Art.  169.)  We  again  remind  the  reader  that  b  will  verify  the 
primitive  equation  (1),  it  being  a  root,  and  it  must  also  verify 
equation  (2)  ;  hence,  b  at  the  same  time  must  verify  the  two 
equations  (1)  and  (2). 

But  if  b  will  verify  equation  (1),  that  equation  is  divisible  by 
(x  —  6),  (Art.  164.),  and  if  it  will  verify  equation  (2),  that  equa- 
tion also,  is  divisible  by  (x  —  b),  and  (x  —  6)  must  be  a  common 
measure  of  the  two  equations  (1)  and  (2).  That  is,  in  case 
the  primitive  equation  has  two  roots  equal  to  b. 

(Art.  170.)  To  determine  whether  any  equation  contains  equal 
roots,  take  its  derived  polynomial  by  the  rule  in  (Art.  168.),  and 
seek  the  greatest  common  divisor  (Art.  27.),  [which  designate  by 


GENERA L  THEORY  OF  EQUATIONS.  ogj 

(/)),]  of  the  given  equation  and  its  first  derived  polynomial ; 
and  if  the  divisor  D  is  of  the  first  degree,  or  of  the  form  of  x — h, 
then  the  equation  has  two  roots  each  equal  to  h. 

If  no  common  measure  can  be  found,  the  equation  contains  no 
equal  roots.  If  D  is  of  the  second  degree,  with  reference  to  x, 
put  Z)=0,  and  resolve  the  equation ;  and  if  D  is  found  to  be  in 
the  form  of  (x — A)2 ;  then  the  given  equation  has  three  roots 
equal  to  h. 

If  D  be  found  of  the  form  of  (x — h)(x — h')t  then  the  given 
equation  has  two  roots  equal  to  h,  and  two  equal  to  h'. 

Let  D  be  of  any  degree  whatever  ;  put  Z)=0,  and,  if  possible, 
completely  resolve  the  equation  ;  and  every  simple  root  of  D  is 
twice  a  root  in  the  given  equation  ;  every  double  root  of  D  will 
be  three  times  a  root  in  the  given  equation,  and  so  on. 

EXAMPLES. 

1.  Does   the   equation   x4 — 2xs — 7x2-}-2Qx — 12=0   contain 
any  equal  roots,  and  if  so,  find  them  ? 

Its  derived  polynomial  is  4x3 — 6x2 — 14#-f20. 

The  common  divisor,  by  (Art.  27.),  is  found  to  be  x — 2  » 
therefore,  the  equation  has  two  roots,  equal  to  2. 

The  equation  can  then  be  divided  twice  by  x — 2,  or  once  by 
(x — 2)2,  or  by  fl2— 4#-{-4.  Performing  the  division,  we  find 
the  quotient  to  be  xz-^-2x — 3,  and  the  original  equation  is  now 
separated  into  the  two  factors, 

(xt—lx+^tf+Zx— 3)  =0. 

The  equation  can  now  be  verified  by  putting  each  of  these 
factors  equal  to  zero.  From  the  first  we  have  already  x=2r 
and  2,  and  from  the  second  we  may  find  x=\  or  —3;  hence, 
the  entire  solution  of  the  equation  gives  1,  2,  2,  — 3  for  the  four 
roots. 

2.  The  equation  arH-2;r4—  lla?— 8x*-{-2Qx+lG=Q  has  two 
equal  roots ;  find  them.  Jlns.     2  and  2. 

3.  Does  the  equation  a?— 2#4+3r} — 1x*+8x— 3=0  contain 
equal  roots,  and  how  many  ? 

Ans.    It  contains  three  equal  roots,  each  equal  to  1 

24 


282  ELEMENTS  OF  ALGEBRA. 

4L  Find  the  equal  roots,  if  any,  of  the  equation 

x3~\-tf  —  16#-f  20=0.  tins*     2  and  2. 

5.  Find  the  equal  roots  of  the  equation 

X4_|_2^3—3x2—  4^+4=0. 
Jlns.     Two  roots  equal  to  1,  and  two  roots  equal  —2. 

6.  Find  the  equal  roots  of  the  equation 

a?—  53H-10a:--8==0. 

JJns.     It  contains  no  equal  roots. 

(Art.  171.)  Equations  which  have  no  commensurable  roots,  or 
those  factors  of  equations  which  remain  after  all  the  commensu- 
rable and  equal  roots  are  taken  away  by  division,  can  be  resol- 
ved only  by  some  method  of  approximation,  if  they  exceed  the 
third  or  fourth  degree.  It  is  possible  to  give  a  direct  solution  in 
cases  of  cubics  and  in  many  cases  of  the  fourth  degree  ;  but,  in 
practice,  approximate  methods  are  less  tedious  and  more  conve- 
nient. 

We  may  transform  any  equation  into  another  whose  roots 
shall  be  greater  or  less  than  the  roots  of  the  given  equation  by 
a  given  quantity. 

Suppose  we  have  the  equation 

a*+Jla*+£a*+Ca*+Dx+JE=0,  .....  (1) 

and  require  another  equation,  whose  roots  shall  be  less  than  those 
of  the  above  by  a. 

Put  x=a-\-y,  and,  of  course,  the  equation  involving  y  will 
have  roots  less  than  that  involving  x,  by  a,  because  y=x  —  «,  or 
?/  is  less  than  x,  by  a. 

In  place  of  xt  in  the  above  equation,  write  its  equal  (a~\-y) 
and  we  have 


By  expanding  and  arranging  the  terms  according  to  the  powers 
of  t/,  we  shall  have,  as  in  (Art.  168.), 


EQUAL  ROOTS. 


E 


(2) 


After  a  little  observation,  these  transformations  may  be  made 
very  expeditiously,  for  the  first  perpendicular  column  may  be 
written  out  by  merely  changing  x  to  a,  in  the  original  equation, 
and  then,  each  horizontal  column  run  out  by  the  law  of  the 
binomial  theorem. 

Thus  a5  becomes  5a4,  and  this,  again,  10a3,  &c. 

Now,  the  first  column  of  the  right  hand  member  of  this  equa- 
tion consists  entirely  of  known  quantities ;  and  the  coefficients 
of  the  different  powers  of  y  are  known  ;  hence  we  have  an 
equation,  involving  the  several  powers  of  y,  in  form  of  equa- 
tion (1), 

Or,  ^+^y+j&y+Cy+/)'y-H5;-==0;  the  equation 
required;  .#',  B',  &c.,  representing  the  known  coefficients  of 
the  different  powers  of  y. 

In  commencing  this  subject,  we  took  an  equation  definite  in 
degree,  for  the  purpose  of  giving  the  pupil  more  definite  ideas ; 
but  it  is  now  proper  to  show  the  form  of  transforming  an  equation 
of  the  most  general  character. 

For  this  purpose,  let  us  take  the  equation 

1 


Now  let  it  be  required  to  transform  this  equation  into  another 
whose  roots  shall  be  less  than  the  roots  of  this  equation  by  a. 

Put  x=a-{-y,  as  before ;  then 


284 


ELEMENTS  OF  ALGEBRA. 

%  .2 

rd      ^ 


CL>      cC 

•+j    9 

.f^f 

rl  ^ 

§  •<; 

'f3    TJ 

eS       **** 

o              .  .           g  j> 

CM   5^ 

1  ? 

,                               .        *     3  2 
%>                                         §  1 

2    ^53 

s  ^ 

I                                         «2 

«u   ?T 

__.                                                                                          r^3     HH 

55          ^N 

«                                                                                                       1 

§    *s                      «"  s 

•        •                                 o    §       «s|§ 

-:    :  :            i!    1 

^^                                                                       ^     i^ 

T 

if 

•i 

1  •§ 

f 

•                                               —  (    t,-.     ^         • 

OD     S 

CO        f+z 

'~-< 

i  •               1  °  I  • 

& 

^3            •                                                              03       C       m            ' 

«       cT       •                               ^  •§    ^     C 

p 

C/)  ep*i 

.f% 

>C    o 

I 

i    I     :                £ts  *  o 

?3            ff                                                                            M       03      O 

7      >:N  :   o 

«M    7                                                         'Jo     «     &       * 

s      1                          i  «  *    • 

^s  J,    -                >.  ^^   : 

heretofore,  p 

!l 

I  i 

0    ^ 

f 

4-       4-     *                              o  ^    »    o 

'                     '              •                                                              03       rt       S3           || 

IB 

C    o 

.2 

o    ^    hn   n^ 

CJ 

8   ^ 

03 

•    III;        s  -^  + 
:    ^  5,  ^            «  -t'l  1 

a 

II 
(i 

S  .| 

•2  .S 
1    2 

*^     c^i     • 

^S    *Ki        ^ 

1 

1    1  11          i|J 

j^  ^T^  j^                       ^    »    ri 

•I 
1 

»:|i| 

55   s   ^ 

S      SH    g, 

C.tf    *? 

03 

1    II               ll'i   = 

*8 

^^^ 

^    0     "^ 

e               S353C3               S3     *>H         ^     ^     PH     n  . 
^^     tt5     C  ^             C**     ^^       ^^     pj     ^i.    F™H 

^*                *           '                                                             c3      c3 

i 

u 

^T*  i3    .2 

3« 

"***     S     <u 
o    ^    ^ 

1 

S     o   "^ 

«*    •.    fc 

"H      "5        S 

.S 

.    «l    64 

p. 

•             f)      ,                 ^ 

u? 

^    "^    "53 

tfi 

"t^      O     +•* 

*.-< 

s    p    ^ 

^j      *-«    ^* 

-*-> 

^~     S   ^^ 

^ 

EQUAL  ROOTS  285 

If  we  should  desire  to  make  the  third  term  (counting  from  the 
highest  power  of  y]  of  equation  (2)  to  disappear,  we  must 

Put  10rt2-H«tf0+-#=0  ;  and  this  involves  the 
solution  of  an  equation  of  the  second  degree,  to  find  the  definite 
value  of  a.  To  make  the  fourth  term  disappear  would  require 
the  solution  of  an  equation  of  the  third  degree  ;  and  so  on. 

If  a  is  really  a  root  of  the  primitive  equation,  then  x=a,  <y=0, 
and  each  perpendicular  column  of  the  transformed  equation  is  0. 

If  we  designate  the  first  perpendicular  column  of  the  general 
transformed  equation  by  X,  and  the  coefficients  of  the  succeeding 
columns  by 

X'     X"       X"' 


The  coefficients  of  the  different  powers  of  y,  as  X',  X",  X'",  &c. 
arc  called  derived  polynomials,  because  each  term  of  X'  can  be 
derived  from  the  corresponding  term  of  X  ;  and  each  term  of  X" 
can  be  derived  from  the  corresponding  terms  of  X',  by  the  law  of 
the  binomial  theorem,  as  observed  in  the  first  part  of  this  article. 
But,  to  recapitulate  : 

X  is  derived  from  the  given  equation  by  simply  changing  x 
to  a. 

X'  is  derived  from  X  by  multiplying  each  of  the  terms  of  X 
by  the  exponent  of  a,  in  that  term,  and  diminishing  that  expo- 
nent by  unity,  and  dividing  by  the  exponent  of  y  increased  by  1. 

X"  is  derived  from  X',  in  the  same  manner  as  X'  is  derived 
from  X  ;  and  so  on. 

X'  is  called  the  first  derived  polynomial  ;  X"  the  second,  &c. 

To  show  the  utility  of  this  theorem,  we  propose  to  transform 
the  following  equations  : 

1.  Transform  the  equation 

x4—l2x3+l7x2—  9*4-7=0, 

into  another,  which  shall  not  contain  the  3d  power  of  the  un- 
known quantity. 


286  ELEMENTS  OF  ALGEBRA. 

By  (Art.  172  ),  put  .r=7/+^  .....   or-  ...  x=3+y 

Here     «=3     and     m=4. 

X  =(3)4—12(3)3+17(3)2—  9(3)-}-7  •  •  .  or  .  .  .  X   =—110 
X'  =4(3)3—  36(3)24-34(3)—  -  9  ......  or  ...  X'  =—123 

~=6(3)2—  36(3)4-17  ..........  or.  .  .  —  =—  37 

z  <& 


Y"' 
12  ..............  m---M 

Therefore  the  transformed  equation  must  be 
2.  Transform  the  equation 


into  another  wanting  its  second  term.     Put  x=2-{-y. 

X  =(2)3—  6(2)2+13(2)—  12  .......  or.  .  .X   =—2 

X'  =3(2)2—  12(2)+13    ..........  or.  -  •  X'  =+1 

6  ...............  or...=     0 


X'"  X' 

21T1  ...................  or---2 

Therefore  the  transformed  equation  must  be 
2/3+ 

3.  Transform  the  equation 


into  another  whose  roots  shall  be  less  by  2. 

Put  a?=24-y.  Result,    ?/44-4?/3— 24y=0, 

As  this  transformed  equation  has  no  term  independent  of  y, 
y=0  will  verify  the  equation ;  and  x=2  will  verify  the  original 
equation,  and,  of  course,  is  a  root  of  that  equation. 


EQUAL  ROOTS.  287 

4.  Transform  the  equation 


into  av.other  whose  roots  shall  be  greater  by  3. 

Put  x=—3-}-y.  Result,    2/4+4?/3-r-9i/2—  42y=0. 

5    Transform  the  equation 

x4—  8z3-{-arJ+82;r—  60=0, 
into  another  wanting  its  second  term. 

Result,    y4—  23*/2+22i/-f-60=0. 

(Art.  173.)  We  may  transform  an  equation  by  division,  as 
well  as  by  substitution,  as  the  following  investigation  will  show. 

Take  the  equation 

*44-^34-  Bx>+Cx+D=Q  ........  (1) 

If  we  put  x=a-}-y,  in  the  above  equation,  it  will  be  trans- 
formed (Art.  D.)  into 


As  x—  a-\-y,  therefore  T/—  #  —  a  ;  and  put  this  value  of  y  in 
equation  (2),  we  have 

(*-«)'+  J^-«)3+^V-«)2+X  '(*-«)+X=0.  .  .(3) 

Now  it  is  manifest  that  equation  (3)  is  identical  with  equation 
(1),  for  we  formed  equation  (2)  by  transforming  equation  (l),and 
from  (2)  to  (3)  we  only  reversed  the  operation. 

Now  we  can  divide  equation  (3),  or  in  fact  equation  (1),  by 
(x  —  a).  and  it  is  obvious  that  the  first  remainder  will  be  X. 

Divide  the  quotient,  thus  obtained,  by  the  same  divisor,  (x  —  a). 
and  the  second  remainder  must  be  X'. 

Divide  the  second  quotient  by  (x  —  a),  and  the  third  remaindei 

nmst  be  —  -. 

m 

X'" 

The  next  remainder  mus',  be  —  ,  <fec.,  <fec.,  according  to  the 

2.3 

degrpe  of  the  equation 


238  ELEMENTS  OF  ALGEBRA. 

Now  if  we  reserve  these  remainders,  it  is  manifest  that  they 
may  form  the  coefficients  of  the  required  transformed  equation ; 
taking  the  last  remainder  for  the  first  coefficient ;  and  so  on,  in 
reverse  order. 

For  illustration,  let  us  take  the  third  example  of  the  last  article. 

01         y=x — 2. 
— 16 


— 2x* — Sx 


16H-32  —4x— 16 

16#-{-32  —  4x+  8 


0=X  — 24=X 


2)a:+2(l 
a:— 2 


X'" 
~ 


Hence  the  transformed  equation  is 


0; 
or,  t/4-f-4?/3  —  24i/=0,  as  before. 

For  a  further  illustration  of  this  method,  we  will  again  operate 
on  the  first  example  of  the  last  article. 


EQUAL  ROOTS. 

—  10a>-39 


—  lOic2—  9x- 


— 39a'-f-     7 
_39.r-|-117 

—110— X.     1st  Remainder. 
;_3U-3— 9^—  1  O.r— 39(0.-— Gx— 28 


—28^—39 


—  123  =  X'.     2d  Remainder. 
2S(x— 3  a>— 3).r— 3(1 

—3x—28  X'" 


__.     3d  Remainder. 


Hence   7/4±()j/3—  3 7/— 123^—110  =  0,  must   be   the  trans- 
formed  equation. 

We  shall  have  a  4th  remainder,  if  we  operate  on  an  equation 
of  the  4th  degree;  a  5th  remainder  with  an  equation  of  the  f>tli 
degree;  and,  in  general,  n  number  of  remainders  with  an  equa- 
tion of  the  nth  degree. 
25 


290  ELEMENTS  OF  ALGEBRA. 

But  to  make  this  method  sufficiently  practical,  the  operator 
must  understand 

SYNTHETIC  DIVISION. 

(Art.  174.)  Multiplication  and  division  are  so  intimately  blended 
that  they  must  be  explained  in  connection.  For  a  particular 
purpose  ve  wish  to  introduce  a  particular  practical  form  of  per- 
forming certain  divisions  ;  and  to  arrive  at  this  end,  we  commence 
with  multiplication. 

Algebraic  quantities,  containing  regular  powers,  may  be 
multiplied  together  by  using  detached  coefficients,  and  annexing 
the  proper  literal  powers  afterwards. 

EXAMPLES. 

1    Multiply     a2  +2^+^     by     a+x. 
Take  the  coefficients.     Thus 
1+2+1 
1  +  1 


1+2+1 
1+2+1 


Product,    «...      1+3+3+1 
By  annexing  the  powers,  we  have 


•J.  Multiply     3t*+xy+y*    by    tf—xy+if. 

As  the  literal  quantities  are  regular,  we  may  take  detached 
coefficients,  thus  : 

1  +  1  +  1 


1  +  1  +  1 
__!_  i— i 

1  +  1  +  1 

Product,.  •  •  •    1+0+1+0+1 


SYNTHETIC  DIVISION.  291 

Hero  the  second  and  fourth  coefficients  are  0  ;  therefore  the 
terms  themselves  will  vanish  ;  and,  annexing  the  powers,  we  shall 
have  for  the  full  product 


3.  Multiply     3,r2—  2.r—  1     by 
3—2—1 

4+2 


12—8—4 

6—4—2 

12—2—8—2 
Product,  .  .  .    12^—2^— Sx— 2. 

.  Multiply     x4 — aar'+flrV — «3#+a4     by 
1— 1  +  1— 1  +  1 


1—1  +  1—  1+  1 
+  1—  1  +  1—1  +  1 

1+0+0+0+0+1 

As  all  the  coefFjcients  are  zero  except  the  first  and  last,  there- 
fore the  product  must  be 


(Art.  175.)  Now  if  we  can  multiply  by  means  of  detached 
coefficients,  in  like  cases  we  can  divide  by  means  of  them. 

Take  the  last  example  in  multiplication,  and  reverse  it,  that  is, 
divide  a^+a5  by  a-+a. 


Here  we  must  suppose  all  the  inferior  powers  of  x5  and  a5 
really  exist  in  the  dividend,  but  disappear  in  consequence  of  their 
coefficients  being  zero  ;  we  therefore  write  all  the  coefficients  of 
the  regular  powers  thus  : 


292  ELEMENTS  OF  ALGEBRA. 

Divisor.  Dividend.  Quotient. 


Annexing  the  regular  powers  to  the  quotient,  we  have  x* — ax3-^- 
azx2 — «3vT+  a4,  for  the  full  quotient. 

2.  Divide   a5— 5a46+lO«362—  10«263+5a&4— tf       by      a8— 
2ab+bz. 

1—2+1)1—5+10—10+5—1(1—3+3—1 
1—2+   1 
—34-9—10 
—34-6—  3 

"   3—7+5 
3_6+3 


—1+2-1 
—1+2—1 

These  coefficients  are  manifestly  the  coefficients  of  a  cube, 
therefore  the  powers  are  readily  supplied,  and  are 


N.  B.  If  we  change  the  signs  of  the  coefficients  in  the  divisor, 
except  the  first,  and  then  add  the  product  of  those  changed  terms, 
we  shall  arrive  at  the  same  result. 

Perform  the  last  example  over  again,  after  changing  the  signs 
of  the  second  and  third  terms  of  the  divisor.  Thus, 


SYNTHETIC  DIVISION.  093 

l-f2__l)l_5_}-10— 10+5— 1(1— 3+  3—1 
l-j-2—  1 

Sum    •  •    *— 3-f-  9—10 
_3_  6+  3 


Sum    .  .  .  .    ;        3 —  7+5 

3-f   6—3 


Sum    .......  *—  J+2—  J 

—  1—2+1 

Sum    .......       ~~*~0+0 

3.  Divide  or3—  Gr'+l  1#—  6     by     x—2. 
Change  the  sign  of  the  second  term  of  the  divisor. 
1+2)1—6+11—6(1—4+3 
1+2 
—4+11 
—4—  8 


3+6 


Let  the  reader  observe,  that  when  the  first  figure  of  the  divisor 
is  1,  the  first  figure  of  the  quotient  will  be  the  same  as  the  first 
figure  of  the  dividend  ;  and  the  succeeding  figures  of  the  quotient 
are  the  same  as  the  first  figures  of  the  partial  dividends. 

Now  this  last  operation  can  be  contracted. 

AY  rite  down  the  figures  of  the  dividend  with  their  proper  signs, 
and  the  second  figure  of  the  divisor,  with  its  sign  changed,  on 
the  right.  Thus 

1—6+1  1—6(2.=  Divisor 
2—  8+6 


(1—44-3)      o 

The  first  figure,  1,  is  brought  down  for  the  first  figure  of 
the  quotient. 

The  divisor,  2,  is  put  under  — 6  ;  their  sum  is  —4,  which, 
multiplied  by  2,  and  the  product  — 8  put  under  the  next  term, 


294  ELEMENTS  OF  ALGEBRA. 

the  sum  of  +11  —  8  is  3,  which  multiplied  by  2,  gives  6,  and  the 
sum  of  the  last  addition  is  0,  which  shows  that  there  is  no 
remainder. 

The  numbers  in  the  lower  line  show  the  quotient,  except  the 
last  ;  that  shows  the  remainder,  if  any. 

This  last  operation  is  called  synthetic  division. 

4.  Divide  a,'3-r-2r!—  -Sx  —  24     by     rr  —  3. 

COMMON  METHOD. 


a-3—  3.r2 


SYNTHETIC  METHOD. 

14-2—  8—24(3 
3+15-1-21 

(14-54-7)—  3 

Now  we  are  prepared  to  work  the  examples  in  (Art.  E.)  in  a 
more  expeditious  manner. 

Transform  again,  the  equation  #4 — 4x3 — 8#4-32=0,  to  an 
other,  whose  roots  shall  be  less  by  2. 

This  equation  has  no  term  containing  a,12,  therefore  the  coefli 
cient  of  a?  must  be  taken  =0,  if  we  use  Synthetic  Division. 

FIRST    OPERATION. 

1—  4±0— 84-32  (2 
2—4—8—32 


(1—2—4—16),  0=X. 

SECOND    OPERATION. 

1—2—4—16(2 
2±0—  8 

(l-f-0— 4),-24=X'. 


SYNTHETIC  DIVISION.  295 

THIRD    OPERATION. 
IrbO — 4  (2 

2+4 

Y" 
(1+2)     0=±-. 

FOURTH    OPERATION. 

1+2(2 
2 


(fence  oui  transformed  equation  is  y4-f-4t/3 — 24*/=0,  as 
before. 

To  transform  an  equation  of  the  fourth  degree,  we  must  have 
four  operations  in  division ;  an  equation  of  the  nth  degree  n 
opeiations,  as  before  observed. 

But  these  operations  may  be  all  blended  in  one.     Thus 

j     __4     ±o     —  8         32     (2 

2—4—8     —32 

0"=:X 


-2 
2 

—4 
0 

—16 

—  8 

=X 

0 
2 

2 
2 

—  4 

+4 

0 

—24 

X" 

=  
2 

2.3* 

We  omit  the  first  column,  except  in  the  first  line,  as  there  aro 
no  operations  with  it. 

The  pupil  should  observe  the  structure  of  this  operation.  It 
is  an  equation  of  the  4th  degree,  and  there  are  four  sums  in  nd- 
dition,  in  the  2d  column;  three  in  the  next;  two  in  the  next, 
&.C.,  giving  the  whole  a  diagonal  shape. 


29 U  ELEMENTS  OF  ALGEBRA. 

Transform  the  equation  x* — l2x3-\-l7x2 — O.r+7  —  0,  into  an- 
other whose  root  shall  be  3  less. 

OPERATION. 

1     —12     +17     —     9     _J-     7     (3 
4.  3     —27     —  30     —117 


_  y     —io     —  39     — 110=X 

-f  3     —18     —  84 


—  6     — 28     —  123=X' 
4.3—9 

-  3     -37=^  , 
3 

^51; 

Hence  the  transformed  equation  is 

10=0. 


Transform  tlie  equation  x3 — l'2x — 28  =  0,  into  another  whose 
roots  shall  be  4  less. 


0 

—12 

—28     ( 

4 

+  10 

+  10 

4 

4 

—  12  =  ] 

4 

32 

IF 

~~36  =  : 

V' 

4 

Hence  the  transformed  equation  must  be     ?/3+12?/s+30i/ — 
12  —  0,  on  the  supposition  that  we  put  y=x — 4. 

Transform  the  equation  #3 — 10a^+3a: — 0946=0,  into  another 
whose  roots  shall  be  less  by  20. 
Put #=20+?/. 


SYNTHETIC  DIVISION.  29? 


—10 

3 

—6946 

(20 

20 

200 

4060 

10 

203 

—2886 

20 
30 

600 

803 

20 

= 

50 


The  three  remainders   are  the  numbers  just  above  the  doulde 
lines,  which  give  the  following  transformed  equation : 
y+50?/2+803?/— 2886=0. 

Transform  this  equation  into   another,  whose  roots  shall  be 
less  by  3.     Put  y=3-\-z. 


50 
3 

53 
3 

803 
159 

962 

168 

—2886 
+2886 

(3 

0 

56 

a 

1130 



59 
Hence  the  second  transformed  equation  is 


This  equation  may  be  verified  by  making  2=0  ;  which  gives 
y—3     and     #=20+3=23. 

Thus  we  have  found  the  exact  root  of  the  original  equation  by 
successive  transformations  ;  and  on  this  principle  we  shall  here- 
after give  a  general  rule  to  approximate  to  incommensuralle  roots 
of  equations  of  any  degree  ;  but  before  the  pupil  can  be  prepared 
to  comprehend  and  surmount  every  difficulty,  he  must  pay  more 
attention  to  general  theory,  as  developed  in  the  following 
Chapter 


298  ELEMENTS  OF  ALGEBRA. 

CHAPTER  III. 

GENERAL  PROPERTIES  OF  EQUATIONS. 

(Art.  170.)  Jlny  equation,  having  only  negative  roots,  will 
have  all  its  signs  positive. 

If  wo  take  —  a,  —  b,  —  c,  &c.,  to  represent  the  roots  of  an 
equation,  the  equation  itself  will  be  the  product  of  the  factors  ; 

(z-r-a),  (x+b),  Or-fc),  &c.,  =0  : 

and  it  is  obvious  that  all  its  signs  must  be  positive. 

From  this,  we  decide  at  once,  that  the  equation  a?4-f-3a^-f- 
6a?+6=0  ;  or  any  other  numeral  equation,  having  all  its  signs 
plus,  can  have  no  rational  positive  roots. 

(Art.  177.)  Surds,  and  imaginary  roots,  enter  equations  by 
pairs. 

Take  any  equation,  as 


and  suppose  (a-f-^/6)  to  be  one  of  its  roots,  then  («  —  Jb)  must 
be  another. 

In  place  of  x,  in  the  equation,  write  its  equal,  and  we  have 


By  expanding  the  powers  of  the  binomial,  we  shall  find  some 
terms  rational  and  some  surd.  The  terms  in  which  the  odd 
powers  of  ,Jb  are  contained  will  be  surd  ;  the  other  terms 
rational  ;  and  if  we  put  R  to  represent  the  rational  part  of  this 
equation,  and  SJb  to  represent  the  surd  part,  then  we  must  have 


But  these  terms  not  having  a  common  factor  throughout,  cannot 
equal  0,  unless  we  have  separately  /?=0,  and  #=0;  and  if  this 
be  the  case  we  may  have 


This  last  equation,  then,  is  one  of  the  results  of 
being  a  root  of  the  equation. 


GENERAL  PROPERTIES  OF  EQUATIONS.  299 

Now  if  we  write  (a  —  Jb)  in  place  of  a?,  in  the  original  equ* 
tion,  and  expand  the  binomials,  using  the  same  notation  as 
before,  we  shall  find 


But  we  have  previously  shown  that  this  equation  must  be  true  ; 
and  any  quantity,  which,  substituted  for  x,  reduces  an  equation 
to  zero,  is  said  to  be  a  root  of  the  equation  ;  therefore  (a  —  Jit) 
is  a  root. 

The  same  demonstration  will  apply  to  (-f~Va)»  (  —  >/a)»  to 


—  «,    —  J  —  a,  and  to  imaginary  roots  in  the  form  of 


(a+bj—l), 

(Art.  178.)  If  we  change  the  signs  of  the  alternate  terms  of 
an  equation,  it  will  change  the  signs  of  all  its  roots. 
At  first,  we  will  take  an  equation  of  an  even  degree. 
If  a  is  a  root  to  the  equation 

x*+J33*+£a*+Cx+D=0  .......  (1) 

then  will  —  a  be  a  root  to  the  equation 

x*—£x*+Bx*—Cx+I)=Q  .......  (2) 

Write  a  for  x,  in  equation  (I),  and  we  have 


Now  write  —  a  for  re,  in  equation  (2),  and  we  have 

a4±rfa*+Ea2-}-Ca-{-I}=()  .......   (4) 

Equations  (3)  and  (4)  are  identical  ;  therefore  if  a,  put  for  .c  in 
equation  (1),  gives  a  true  result,  —  a  put  for  x  in  equation  (2), 
gives  a  result  equally  true. 

We  will  now  take  an  equation  of  an  odd  degree. 

If  the  equation          xs-\-^x2-\-Bx-\-C=Ot 
have  a  for  a  root,  then  will  the  equation 


have  —a  for  a  root. 
From  the  first 
From  the  second  —a3  —  4#a2—  /?a—  6V.-0. 


300  ELEMENTS  OF  ALGEBRA. 

This  second  equation  is  identical  with  the  first,  if  \ve  change 
all  its  signs,  which  does  not  essentially  change  an  equation. 

The  equation  x4-f  a:3—  l9x2-{-l  l;r-{-30=0,  has  —1,  2,  3, 
and  —-5,  for  its  roots  ;  then  from  the  preceding  investigation 
\vc  learn  that  the  equation 


must  have  1  —  2,  —  3,  and  +5  for  its  roots. 

(Art.  179.)  If  we  introduce  one  positive  root  into  an  equa- 
tion, it  will  produce  at  least  one  variation  in  the  signs  of  its 
term;  if  two  positive  roots,  at  least  two  variations. 

The  equation  x2-\-x-\-l.=Q,  having  no  variation  of  signs,  can 
have  no  positive  roots.  (Art.  176.)  Now  if  we  introduce  the 
root  +2,  or  which  is  the  same  thing,  multiply  by  the  factor  x  —  2, 

x*+  a?+l 
x  —  2 


x--  x       x 
—Zx2— 2x— 2 

Then a;3-— a*-— a:— 2=0; 

and  here  we  find  one  variation  of  signs  from  -\-x*  to  — £2,  and 
one  permanence  of  signs  through  the  rest  of  the  equation, 

If  we  take  this  last  equation  and  introduce  another  positive 
root,  say  -|-5,  or  multiply  it  by  x — 5,  we  shall  then  have 


1—1    —1    —2 

—5    -[-5    +5  +10 


Here  are  two  variations  of  signs,  one  from  4-a?4  to  —  O.r3,  and 
another  from  —  Gx3  to  -f-4;r2. 

And  thus  we  might  continue  to  show  that  every  positive  root, 
introduced  into  an  equation,  will  produce  at  least  one  variation 
of  signs.  But  we  must  not  conclude  that  the  converse  of  this 
proposition  is  true. 


GENERAL  PROPERTIES  OF  EQUATIONS.  3QJ 

Every  positive  root  will  give  one  variation  of  signs  ;  but 
every  variation  of  signs  does  not  necessarily  show  the  existence 
of  a  positive  root. 

For  an  equation  may  have 

(a+bj—V),    («—  V^f),    —  c,    —d, 

for  roots  ;  then  the  equation  will  be  expressed  by  the  product  of 
the  factors 

(a*—  2ax+az+b*)  (x+c)  (x+d)=Q. 

As  one  of  these  terms,  (  —  2ax),  has  the  minus  sign,  it  will 
produce  some  minus  terms  in  the  product  ;  and  there  must  neces- 
sarily be  variations  of  signs  ;  yet  there  is  no  positive  root.  At 
the  same  time,  the  whole  factor  in  which  the  minus  term  is  found, 
must  be  plus,  whatever  value  be  given  to  x,  as  it  is  evidently 
equal  to  (x  —  a)2-\-b*,  the  sum  of  two  squares. 

The  equation 

^  —  2x3—x2-\-2  x  +10=0, 

has  two  variations  of  signs,  and  two  permanences,  but  the  roots 
are  all  imaginary,  viz., 

T    and    -.1-=T. 


If  it  were  not  for  imaginary  roots,  the  number  of  variations 
among  the  signs  of  an  equation  would  indicate  the  number  of 
plus  roots  :  and  this  number,  taken  from  the  degree  of  the  equa- 
tion, would  leave  the  number  of  negative  roots;  or  the  number 
of  permanences  of  signs  would  at  once  show  the  number  of 
negative  roots. 

To  determine  a  priori  the  number  of  real  roots  contained  in 
any  equation,  has  long  bafHed  the  investigations  of  mathemati- 
cians ;  and  the  difficulty  was  not  entirely  overcome  until  1829, 
when  M.  Sturm  sent  a  complete  solution  to  the  French  Academy. 
The  investigation  is  known  as  Sturm's  Theorem,  and  will  be  pre- 
sented in  the  following  Chapter. 

LIMITS    TO    ROOTS. 

(Art.  180.)  All  positive  roots  of  an  equation  are  comprised 
between  zero  and  infinity  ;  and  all  negative  roots  between  zera 


302  ELEMENTS  OF  ALGEBRA. 

and  minus  infinity  ;  but  it  is  important  to  be  able  at  once  to 
assign  much  narrower  limits. 

We  have  seen,  (Art.  179.),  that  every  equation,  having  a  posi- 
tive root,  must  have  at  least  one  variation  among  its  signs,  and 
at  least  one  minus  sign. 

If  the  highest  power  is  minus,  change  all  the  signs  in  the 
equation. 

Now  we  propose  to  show  that  the  greatest  positive  root  must 
be  less  than  the  greatest  negative  coefficient  plus  one. 

Take  the  equation 

=Q. 


It  is  evident,  that  as  the  first  term  must  be  positive  for  all  de- 
grees, x  must  be  greater  and  greater,  as  more  of  the  other  terms 
are  minus  :  then  x  must  be  greatest  of  all  when  all  the  othei 
terms  are  minus,  and  each  equal  to  the  greatest  coefficient,  (D 
being  considered  the  coefficient  of  a?°). 

Now  as  A,  B,  &c.,  are  supposed  equal,  and  all  minus,  we  shall 
have 


For  the  first  trial  take  x=J2,  and  transpose  the  minus  quantity, 
and  we  have 


Divide  by  ^4,  and  we  have 

1=1+xh/F*~;#~3* 

Now  we  perceive  that  the  second  member  of  the  equation  is 
greater  than  the  first,  and  the  expression  is  not,  in  fact,  an  equa- 
tion. x=rf  proves  x  not  to  be  large  enough. 

For  a  second  tiial  put  x= 

Then 


Dividing  by  (^-fl)'1,  we  have 

A  Jl  A  A 

-*  8        -4' 


We  retain  the  sign  of  equality  for  convenience,  though  the 


GENERAL  PROPERTIES  OF  EQUATIONS.  303 

members  are  not  equal.  The  second  member  consists  of  terms 
in  geometrical  progression,  and  their  sum,  (Art.  120),  is 

1 — T~aTT~\\'     Hence  tne  first  member  is  greater  than  the  second, 

which  shows  that  (.#+1)  substituted  for  x,  is  too  great.  But  Jl 
was  too  small,  therefore  the  real  value  of  x,  in  the  case  under 
consideration,  must  be  more  than  Jl  and  less  than  (.#+1). 

77m/  is,  the  greatest  positive  root  of  an  equation,  in  the 
most  extreme  case,  must  be  less  than  the  greatest  negative 
coefficient  plus  one. 

In  common  cases  the  limit  is  much  less. 

From  this,  we  at  once  decide  that  the  greatest  positive  root  of 
the  equation  x5 — S-r'+To:3 — 8xz — 9a? — 12=0,  is  less  than  13. 

Now  change  the  second,  and  every  alternate  sign,  and  we 
have  the  equation 

ar5+3.r4+7r3+8;r2— 9^+12=0. 

The  greatest  positive  root,  in  this  equation,  is  less  than  10  ;  but, 
by  (Art.  178.),  the  greatest  positive  root  of  this  equation  is  the 
greatest  negative  root  of  the  preceding  equation;  therefore  10  is 
the  greatest  limit  of  the  negative  roots  of  the  first  equation  ;  and 
all  its  roots  must  be  comprised  between  +13  and  —10  ;  but  as 
this  equation  does  not  present  an  extreme  case,  the  coefficients 
after  the  first  are  not  all  minus,  nor  equal  to  each  other ;  there- 
fore the  real  limits  of  its  roots  must  be  much  within  +13  and 
-—10.  In  fact,  the  greatest  positive  root  is  between  3  and  4,  and 
the  greatest  negative  root  less  than  1 . 

If  it  were  desirable  to  find  the  limits  of  the  least  root,  put 

0,=-,  and  transform  the  equation  accordingly.     Then  find,  as 

y 

just  directed,  the  greatest  limit  of  y,  in  its  equation  ;  which  will, 
of  course,  correspond  to  the  least  value  of  x  in  its  equation. 

(Art.  181.)  If  we  substitute  any  number  less  than  the  least 
root,  for  the  unknown  quantity,  in  any  equation  of  an  even 
degree,  THE  RESULT  WILL  BE  POSITIVE.  *ftnd  if  the  degree  of  the 
equation  be  odd,  THE  RESULT  WILL  BE  NEGATIVE. 

Let  «,  b,  c,  &c.,  be  roots  of  an  equation,  and  x  the  unknown 


3J4  ELEMENTS  OF  ALGEBRA. 

quantity.     Ayso,  conceive  a  to   be   the   least  root,  /;   the  next 
greater,  and  so  on.     Then  the  equation  will  be  represented  by 
(x — a)(x — b)(x — c)(a? — rf),  &c.,  =0. 

Now  in  the  place  of  x  substitute  any  number  h  less  than  a,  and 
the  above  factors  will  become 

(h—a)(h—b)(h—c}(h—d),  &c. 

Each  factor  essentially  negative,  and  the  product  of  an  even 
number  of  negative  factors,  is  positive ;  and  the  product  of  an 
odd  number  is  negative ;  therefore  our  proposition  is  proved. 

Scholium. — If  we  conceive  h  to  increase  continuously,  until  it 
becomes  equal  to  a,  the  first  factor  will  be  zero ;  and  the  product 
of  them  all,  whether  odd  or  even,  will  be  zero,  and  the  equa- 
tion will  be  zero,  as  it  should  be  when  A  becomes  a  root. 

If  h  increases  and  becomes  greater  than  a,  without  being 
iqual  to  I),  the  result  of  substituting  it  for  x  will  be  NEGATIVE, 
in  an  equation  of  an  even  degree,  and  positive  in  an  equation 
of  an  odd  degree. 

For  in  that  case  the  first  factor  will  be  positive,  and  all  the 
other  factors  negative ;  and,  of  course,  the  signs  of  their  product 
will  be  alternately  minus  and  plus,  according  as  an  even  or  odd 
number  of  them  are  taken. 

If  h  is  conceived  to  increase  until  it  is  equal  to  b,  then  the 
second  factor  is  zero,  and  its  substitution  for  x  will  verify  the 
equation.  If  h  becomes  greater  than  b,  and  not  equal  to  c,  then 
the  first  two  factors  will  be  positive  ;  the  rest  negative ;  and  the 
result  of  substituting  h  for  x  will  give  a  positive  or  negative 
result,  according  as  the  degree  of  the  equation  is  even  or  odd. 

If  we  conceive  h  to  become  greater  than  the  greatest  root, 
then  all  the  factors  will  be  positive,  and,  of  course,  their  product 
positive. 

For  example,  let  us  form  an  equation  with  the  four  roots  — 5, 
2,  6,  8,  and  then  the  equation  will  be 

(^+5)(x— 2)  (a?— 6)(ar— 8)=0, 
Or.  ...  #4— 11^— 4z2+284z— 480=0. 
(The  greater  a  negative  number  is,  the  less  it  is   considered.) 


GENERAL  PROPERTIES  OF  EQUATIONS.  305 

Now  if  we  substitute  — 6  for  x,  in  the  equation,  the  result  must 
be  positive.  Let  — 6  increase  to  — 5,  and  tho  result  will  be  0. 
Let  it  still  increase,  and  the  result  will  be  negative,  until  it  has 
increased  to  +2,  at  which  point  the  result  will  again  be  0. 

If  we  substitute  a  number  greater  than  2,  and  less  than  6,  for 
#,  in  the  equation,  the  result  will  again  be  positive.  A  number 
between  6  and  8,  put  for  x,  will  render  the  equation  negative  ; 
and  a  number  more  than  8  will  render  the  equation  positive  ; 
and  if  the  number  is  still  conceived  to  increase,  there  will  be  no 
more  change  of  signs,  because  we  have  passed  all  the  roots. 

If  in  any  equation  we  substitute  numbers  for  the  unknown 
quantity,  which  differ  from  each  other  by  a  less  number  than 
the  difference  between  any  two  roots,  and  commence  with  a 
number  less  than  the  least  root,  and  continue  to  a  number  greater 
than  the  greatest  root,  we  shall  have  as  many  changes  of  signs 
in  the  results  of  the  substitution  as  the  equation  has  real  roots. 

If  one  real  root  lies  between  two  numbers  substituted  for  the 
unknown  quantity,  in  any  equation,  the  results  will  necessarily 
show  a  change  of  signs. 

If  one,  or  three,  or  any  odd  number  of  roots,  lie  between  the 
two  numbers  substituted,  the  results  will  show  a  change  of  signs 

If  an  even  number  of  roots  lie  between  the  two  numbers  sub- 
stituted, the  results  will  show  no  change  of  signs. 

In  the  last  equation,  if  we  substitute  — 6  for  a?,  the  result 
will  be  plus. 

If  we  substitute  -J-3,  the  result  will  also  be  plus,  and  give  no 
indication  of  the  two  root?  — 5  and  -f-2,  which  lie  between. 

(Art.  182.)  If  an  equation  contains  imaginary  roots,  the 
factors  pertaining  to  such  roots  will  be  either  in  the  form  of 
(a«-f-«)»  or  in  the  form  of  [(a? — a)2-{-&2],  both  positive,  whatever 
numbers  may  be  substituted  for  x,  either  positive  or  negative ; 
hence,  if  no  other  than  imaginary  roots  enter  the  equation,  all 
substitutions  for  x  will  give  positive  results,  and  of  course,  no 
changes  of  sign.  It  is  only  when  the  substitutions  for  x  pass 
real  roots  that  we  shall  find  a  change  of  signs. 

26 


306  ELEMENTS  OF  ALGEBRA. 

CHAPTER  IV. 
GENERAL  PROPERTIES  OF  EQUATIONS—  CONTINUED. 

(Art.  183.)  If  we  take  any  equation  which  has  all  its  roots 
real  and  unequal,  and  make  an  equation  of  its  first  derived  poly- 
nomial, the  least  root  of  this  derived  equation  will  be  greater  than 
the  least  root  of  the  primitive  equation,  and  less  than  the  next 
greater. 

If  the  primitive  equation  have  equal  roots,  the  same  root  will 
verify  the  derived  equation.* 

We  sha)l  form  our  equations  from  known  positive  roots. 

Let  <i,  b,  c,  d,  &c.,  represent  roots  ;  and  suppose  a  less  than 
&,  b  less  than  c,  <fec.,  and  x  the  unknown  quantity.  An  equation 
of  the  second  degree  is 

x2  —  ( 

Its  first  derived  polynomial  is 


If  we  make  an  equation  of  this,  that  is,  put  it  equal  to  0,  we 
shall  have 


Now  if  b  is  greater  than  «,  the  value  of  x  is  more  than  a, 
and  less  than  6,  and  proves  our  proposition  for  all  equations  of 
the  second  degree.  If  we  suppose  a=6,  then  x=a,  in  both 
equations. 

An  equation  of  the  third  degree  is 

^—(a+b+c^+^b+ac+bclx—abc^  .....  (i) 
Its  first  derived  polynomial  is 


This  equation,  being  of  the  second  degree,  has  two  roots,  and 
only  two. 

•  To  ensure  perspicuity  and  avoid  too  abstruse  generality,  we  ope-ate  on 
equations  definite  in  degree  ;  the  result  will  be  equally  satisfactory  to  the 
learner,  and  occupy,  comparatively,  but  little  space. 


GENERAL  PROPERTIES  OF  EQUATIONS.  30? 

Now  if  we  can  find  a  quantity  which,  put  for  x,  will  verify 
equation  (2),  that  quantity  must  be  one  of  its  roots.  If  we  try 
two  quantities,  and  find  a  change  of  signs  in  the  results,  we  are 
sure  a  root  lies  between  such  quantities.  (Art.  181.)  Therefore 
we  will  try  a,  or  write  a  in  place  of  x.  As  b  and  c  are  each 
greater  than  a,  we  will  suppose  that 


With  these  substitutions,  equation  (2)  becomes 


Reduced,  gives  ......   -j-/j/i'=0. 

Therefore  a  cannot  be  a  root;  if  it  were  we  should  have  0=0. 
If  we  now  make  a  substitution  of  b  for  a*,  or  rather  (a-\-li) 
for  a*,  and  reduce  the  equation,  we  shall  find 


It  is  apparent  that  this  quantity  is  essentially  minus,  as  h' 
is  greater  than  h.  Hence,  as  substituting  a  for  a?,  in  the  equa- 
tion, gives  a  small  plus  quantity,  and  b  for  x  gives  a  small 
minus  quantity,  therefore  one  value  of  a?,  to  verify  equation  (2), 
must  lie  between  a  and  b. 

This  proves  the  proposition  for  equations  of  the  third  degree  : 
and  in  this  manner  we  may  prove  it  for  any  degree;  but  the  labor 
of  substituting  for  a  high  equation  would  be  very  tedious. 

If  we  suppose  «=6,  and  put  c=a-\-h',  and  then  substitute  a 
in  place  of  a?,  we  shall  find  equations  (1)  and  (2)  will  be  verified. 

Therefore  in  the  case  of  equal  roots,  the  equation  and  its 
first  derived  polynomial  will  have  a  common  measure,  as  before 
shown  in  (Art.  168). 

If  all  the  roots  of  an  equation  are  equal,  the  equation  itself 
i:.5ay  be  expressed  in  the  form  of 

(x—  a)m=0. 

Its  first  derived  polynomial,  put  into  an  equation,  will  be 
m(x  —  a)m~'=0. 

It  is  apparent  that  the  primitive  equation  has  m  roots  equal  to 
a;  and  the  derived  equation,  (ra—  1),  roots  also  equal  to  a. 


308  ELEMENTS  OF  ALGEBRA. 

Lastly,  take  a  general  equation,  as 


Its  first  derived  polynomial,  taken  for  an  equation,  will  be 

mxm-l+(m—  l)rfxm~*  .....  7?=0. 

We  may  suppose  this  general  equation  composed  of  the 
factors 

(x  —  a)(x  —  b)  (x  —  c),  &c.,  =0, 

and  also  suppose  b  greater,  but  insensibly  greater,  than  a  ;  c 
insensibly  greater  than  b,  &c.  Then  the  equation  will  be  nearly 

(x  —  «)OT=0  ; 
and  its  derived  polynomial, 

m(x  —  «)"l~1=0, 

cannot  have  a  root  less  than  (a),  the  least  root  of  the  primitive 
equation  ;  but  its  root  cannot  equal  a,  unless  the  primitive  equa- 
tion have  equal  roots  ;  therefore  it  must  be  greater. 

By  the  same  mode  of  reasoning  we  can  show  that  the  greatest 
root  of  an  equation  is  greater  than  the  greatest  root  of  its  derived 
equation  ;  hence  the  roots  of  the  derived  equation  are  interme- 
diate, in  value,  to  the  roots  of  the  primitive  equation,  or  contained 
within  narrower  limits. 

(Art.  184.)  If  we.  take  any  equation,  not  having  equal  roots, 
and  consider  its  first  derived  polynomial  also  an  equation, 
and  then  substitute  any  quantity  less  than  the  least  root  of 
either  equation,  for  the  unknown  quantity,  the  result  of  such 
mbstilution  will  necessarily  give  opposite  signs. 

Let  a,  b,  c,  &c.,  represent  the  roots  of  a  primitive  equation, 
and  a',  b',  &c.,  roots  of  its  first  derived  polynomial  ;  x  the  un- 
known quantity.  Then  the  equation  will  be 

(x  —  a)(x  —  b)(x  —  c),  &c.,  to  m  factors  =0  ; 
the  derived  equation  will  be 

(x  —  a')(x  —  b')(x  —  c'),  &c.,  to  (m  —  1)  factors  =0. 

Now  if  we  substitute  h  for  x,  and  suppose  h  less  than  either 
root,  then  every  factor,  in  both  equations,  will  be  negative. 

The  product  of  an  even  number  of  negative  factors  is  positive 
and  the  product  of  an  odd  number  is  negative  ;  and  if  the  factor?, 


GENERAL  THEORY  OF  EQUATIONS       309 

in  the  primitive  equation  are  even,  those  in  the  der  ved  equation 
must  be  odd. 

Hence  any  quantity  less  than  any  root  of  either  equation,  will 
necessarily  give  to  these  functions  opposite  signs. 

(Art.  185.)  Now  if  we  conceive  h  to  increase  until  it  becomes 
equal  to  a,  the  least  root,  the  factor  (x — «)  will  be  0,  and  reduce 
the  whole  equation  to  0.  Let  h  still  increase  and  become 
greater  than  a,  and  not  equal  to  a',  (which  is  necessarily  greater 
than  a,  (Art.  184.),  and  the  factor  (x — a)  will  become  plus,  while 
till  the  other  factors,  in  both  equations,  will  be  minus,  and,  of 
course,  leave  the  same  number  of  minus  factors  in  both  functions, 
which  must  give  them  the  same  sign.  Consequently,  in  passing 
the  least  root  of  the  primitive  equation  a  variation  is  changed  into 
a  permanence. 

Sturm's  Theorem. 

(Art.  186.)  If  we  take  any  equation  not  having  equal  roots,  and  its 
first  derived  polynomial,  and  operate  with  these  functions  as  though 
their  common  measure  was  desired,  reserving  the  several  remain- 
ders with  their  signs  changed,  and  make  equations  of  these  func- 
tions, namely,  the  primitive  equation,  its  first  derived  polynomial 
and  the  several  remainders  with  their  signs  changed,  and  then 
substitute  any  assumed  quantity,  h,  for  x,  in  the  several  functions, 
noting  the  variation  of  signs  in  the  result ;  afterwards  substitute 
another  quantity,  h',  for  x,  and  again  note  the  variation  of  signs ; 
the  difference  in  the  number  of  variation  of  signs,  resulting 
from  the  two  substitutions,  will  give  the  number  of  real  roots 
between  the  limits  h  and  h'. 

If  * — QQ  and  +00  are  taken  for  h  and  h',  we  shall  have  the 
whole  number  of  real  roots  ;  which  number,  subtracted  from  the 
degree  of  the  equation,  will  give  the  number  of  imaginary  roots. 

DEMONSTRATION. 

Let  X  represent  an  equation,  and  X'  its  first  derived  polyno- 
mial. 

In  operating  as  for  common  measure,  denote  the  several  quo- 

*  Symbols  of  infinity. 


310  ELEMENTS  OF  ALGEBRA. 

tients  by  Q,  Q'  Q",  &c.,  and  the  several  remainders,  with  their 
sign$  changed,  by  R,  R',  R",  &c. 

In  these  operations,  be  careful  not  to  strike  out  or  introduce 
minus  factors,  as  they  change  the  signs  of  the  terms  ;  then  a 
re-change  of  signs  in  the  remainder  would  be  erroneous. 

From  the  manner  of  deriving  these  functions,  we  must  have 
the  following  equations  • 

X  R 


R  R 


IF*     -IF 

<fcc.         &c. 
Clearing  these  equations  of  fractions,  we  have 

X  =»X'  Q  — R 

X'=RQ'  — R' 
R  =nR'Q"  — R" 
R'=R"Q"'— R'" 


As  the  equation  X=0  must  have  no  equal  roots,  the  functions 
X  and  X'  can  have  no  common  measure  (Art.  168.),  and  we  shall 
arrive  at  a  final  remainder,  independent  of  the  unknown  quantity, 
and  not  zero. 

Proposition  1.  No  two  consecutive  functions,  in  equations  (A), 
can  become  zero  at  the  same  time. 

For,  if  possible,  let  such  a  value  of  h  be  substituted  for  x,  as 
to  render  both  X'  and  R  zero  at  the  same  time ;  then  the  second 
equation  of  (A)  will  give  R'=0.  Tracing  the  equations,  we  must 
finally  have  the  last  lemainder  RTO  =0;  but  this  is  inadmis- 
sible ;  therefore  the  proposition  is  proved. 

Prop.  2.  If  lien  one  of  the  functions  becomes  zero,  by  giving 


GENERAL  PROPERTIES  OF  EQUATIONS.  311 

a  particular  value  to  x,  the  adjacent  functions  between  which  it 
is  placed  must  have  opposite  signs. 

Suppose  R'  in  the  third  equation,  (A),  to  become  0,  then  the 
equation  still  existing,  we  must  have  R= — R". 

The  truth  of  Sturm's  Theorem  rests  on  the  facts  demonstrated 
in  Arts.  184,  185  and  in  the  two  foregoing  propositions. 

If  we  put  the  functions  X,  X',  R,  R',  &c.,  each  equal  to  0  ; 
that  is,  make  equations  of  them,  and  afterwards  substitute  any 
quantity  for  #,  in  these  functions,  less  than  any  root,  the  firs 
and  second  functions,  X  and  X',  will  have  opposite  signs,  (Art. 
184.) ;  and  the  last  function  will  have  a  sign  independent  of  x, 
and,  of  course,  invariable  for  all  changes  in  that  quantity.  The 
other  functions  may  have  either  plus  or  minuSj  and  the  signs 
have  a  certain  number  of  variations. 

Now  all  changes  in  the  number  of  these  variations  must 
come  through  the  variations  of  the  signs  in  the  primitive  func- 
tion X.  A  change  of  sign  in  any  other  function  will  produce 
no  change  in  the  number  of  variations  in  the  series. 

For,  conceive  the  following  equations  to  exist: 


(B) 


Now  take  x=h,  yet  h  really  less  than  any  root  of  the  equa- 
tions, (B),  and  we  may  have  the  following  series  of  signs : 

X   =  —  1 

X'  =  4- 

R  =  — 
R'  =  — 
R"=  — 
R'"=  4- 

Or  we  mav  have  any  other  order  of  signs,  restricted  only  to  the 
fact  that  the  signs  of  the  two  first  functions  must  be  opposite,  and 
the  last  invariable,  or  unaffected  by  all  future  substitutions. 
are  three  variations  of  signs. 


312  ELEMENTS  OF  ALGEBRA. 

Now  conceive  h  to  increase.  No  change  of  signs  can  take 
place  in  any  of  the  equations,  unless  h  becomes  equal  and  passes 
a  root  of  that  equation  ;  and  as  there  are  no  equal  roots,  no  two 
of  these  functions  can  become  0  at  the  same  time  (Prop.  1.)  ; 
hence  a  change  of  sign  of  one  function  does  not  permit  a  change 
in  another ;  therefore  by  the  increase  of  A,  one  of  the  functions, 
(C),  will  become  0,  and  a  further  increase  of  h  will  change  its 
sign. 

In  the  series  of  signs  as  here  represented,  X'  cannot  be  the 
first  to  change  signs,  for  that  would  leave  the  adjacent  functions, 
X  and  R,  of  the  same  sign,  contrary  to  Prop.  2 ;  nor  can  the 
function  R'  be  the  first  to  change  sign,  for  the  same  reason. 

Hence  X  or  R  or  R"  must  be  first  to  change  sign. 

If  we  suppose  X  to  change  sign,  the  other  signs  remaining  the 
same,  the  number  of  variations  of  signs  is  reduced  by  unity. 

If  R  or  R"  change  sign,  the  number  of  variations  cannot  be 
changed ;  a  permanence  may  be  made  or  reduced,  and  all  cases 
that  can  happen  with  three  consecutive  functions  may  be  ex- 
pressed by  the  following  combinations  of  signs  ; 

+     ±     —    ) 
Or  —     ±     +    j 

either  of  which  gives  one  variation  and  one  permanence. 

Now  as  no  increase  or  decrease  in  the  number  of  variations  of 
signs  can  be  produced  by  any  of  the  functions  changing  signs, 
except  the  first,  and  as  that  changes  as  many  times  as  it  has  real 
roots,  therefore  the  changes  in  the  number  of  variations  of  signs 
show  the  number  of  real  roots  comprised  between  h  and  h'. 

If  h  and  h'  are  taken  at  once  at  the  widest  limits  of  possibility, 
from  —  infinity  to  -1-  infinity,  the  number  of  variations  of  signs 
will  indicate  the  number  of  real  roots;  — and  this  number, 
taken  from  the  degree  of  the  equation,  will  give  the  number  of 
the  imaginary  roots. 

(Art.  187.)  The  foregoing  is  a  full  theoretical  demonstration 
of  the  theorem  ;  but  the  subject  itself  being  a  little  abstruse, 
some  minds  may  require  the  following  practical  elucidation. 


GENERAL  PROPERTIES  OF  EQUATIONS.  3J3 

Form  an  equation  with  the  four  assumed  roots,  1,  3,  4,  6. 

The  equation  will  be 


or  X  =     a?4—  14#3+67#2—  126aH-72=0  Roots  1,  3,  4,  6. 
X  =  4xs—  42r4-J34#—  126  =0.  .  Jfoofa  2,  3.3,  5  nearly. 
R  =13y?—9lx  +153  .........  Moots  2.8,  4.1  nearly. 

R'=72#—  252  .............  Hoot   3.5 

R"=  + 

\ 

Let  the  pupil  observe  these  functions,  and  their  roots,  and  see 
that  they  correspond  with  theory.  The  least  root  of  X  is  less 
than  the  least  root  of  X'.  (Art.  183.)  The  roots  of  any  func- 
tion are  intermediate  between  the  roots  of  the  adjacent  functions. 
This  corresponds  with  (Prop.  2.)  ;  for  if  three  consecutive  func- 
tions have  the  same  sign  as  —  ,  —  -,  —  ,  or  -f-,  +,  -K  the  middle 
one  cannot  change  first  and  correspond  to  (Prop.  2.)  ;  but  signs 
change  only  by  the  increasing  quantity  passing  a  root,  and  it  must 
pass  a  root  of  one  of  the  extreme  functions  first  ;  therefore  the  roots 
of  X'  must  be  intermediate  in  value  between  the  roots  of  X  and 
R  ;  and  the  roots  of  R  intermediate  in  value  between  the  roots 
of  X'  and  R'  ;  and  so  on.  But  the  roots  of  X'  are  within  nar 
rower  limits  than  the  roots  of  X  (Art.  183.)  ;  therefore  the  roots 
of  all  the  functions  are  within  the  limits  of  the  roots  of  X. 

We  will  now  trace  all  the  changes  of  signs  in  passing  all  the 
roots  of  all  the  functions. 

We  will  first  suppose  x  or  h  =  0  ;  which  is  less  than  any 
root  ;  then  as  we  increase  h  above  any  root,  we  must  change  the 
sign  of  that  function,  and  that  sign  only 

We  represent  these  changes  thus  : 

27 


314  ELEMENTS  OF  ALGEBRA. 

X  X'  R  R'  R" 

*  When   x—Q       +  —  +  —  -{-....  4  variations 

#=1.1    — *  —  4-  —  4-.  .  .  .3         « 

"        #=2.1   —  4-*  4.  —  4-  ....  3         " 

"       #=2.9  —  4-  — *  —  _)_....  3        « 

"       #=3.1    4-*  +  —  —  +  ....  2 

»        #=3.4   4"  — *     — —  4"  •  •  •  •  2         " 

"       #=3.7    4-  —  —  -p  _{-....  2 

"       #=4.1    — *  —  —  +  4. .  .  .  .  i         « 

"       #=4.11 —  —  4-*  4-  4-....1         " 

«      #=5.i  —     4-*    4-      4-     -i- ....  i       » 
«      #=6.1  4-*    4-      4-      4-      4 o       « 

We  commenced  with  4  variations  of  signs,  and  end  with  0 
variations,  after  we  have  passed  all  the  roots ;  therefore  the  real 
roots,  in  the  primitive  equation,  must  be  4 — 0=4. 

By  this  it  can  be  clearly  seen,  by  inspection,  that  the  changes 
of  sign  in  all  the  functions,  except  the  first,  produce  no  change 
in  the  number  of  variations. 

In  making  use  of  this  theorem  we  do  not  go  through  the  inter- 
mediate steps,  unless  we  wish  to  learn  the  locality  of  the  roots  as 
well  as  their  number.  We  may  discover  their  number  by  sub- 
stituting a  number  for  x  less  than  any  root,  and  then  one  greater ; 
tne  difference  of  the  variations  of  signs  will  be  equal  to  the  num- 
ber of  real  roots. 

If  we  take  — oo  and  4-QD,  the  sign  of  any  whole  function 
will  be  the  same  as  that  of  its  first  term. 

*  In  making  this  table,  we  did  not  really  substitute  the  numbers  assumed  for 
x,  as  we  previously  determined  the  roots ;  and  as  passing  any  root  changes  the 
sign  in  that  function,  we  write  a  star  against  that  sign  which  has  just 
changed. 


APPLICATION  OF  STURM'S  THEOREM.  315 

CHAPTER  V. 

APPLICATION  OF  STURM'S  THEOREM. 
(Art.  188.)  In  preparing  the  functions,  remember  that  we  are 
at  liberty  to  suppress  positive  numeral  factors. 

EXAMPLES. 

!•  How  many  real  roots  has  the  equation   x3-}-9x=Q  ? 
Here  X  = 

X'= 
R=—  ar+1 
R'=— 
Now  for  x  substitute  — co  or  — 100000,  and  we  see  at  once 

^  X        X'        R        R' 

—        -j-        ~J-       —        2  variations. 

Again,  for   x  put  -f-co  or  +100000,  and  the  resulting  signs 

must  be  ,  , 

•f-        +        —        —        1  variation. 

Hence  the  above  equation  has  but  one  real  root ;  and,  of  course, 
two  imaginary  roots. 

To  find  a  near  locality  of  this  root,  suppose  x=Q,  and  the 

signs  will  be  , 

—        4-         =t        —        2  variations. 

x=l  -|-         4~         rb        —      .1  variation. 

Hence  the  real  root  is  between  0  and  1 

Now  as  we  have  found  x,  in  the  equation  x3-}-Qx — 6=0,  to 
be  less  than  1,  Xs  may  be  disregarded,  and  9x — 6=0,  will  give 
us  the  first  approximate  value  of  a?;  that  is,  a?=.6,  nearly. 

2.  How  many  real  roots  has  the  equation  a;4 — 3xz — 4=0  ? 
X=  a;4— Zx2— 4 
X'=4a:3—  Qx 
R=+25 

If  X——CQ          -f-         —         +2  variations. 

#=-1-03          -|-         -}-         +     0  variation. 
Hence  there  must  be  2  real  roots, and  2  imaginary  roots. 


316  ELEMENTS  OF  ALGEBRA. 

3.  How  many  real  roots  has  the  equation   x"  —  4#3—  621=0  1 
(See  Art.  103.) 

X=z6—  4^—621 

X'=x5—  23? 

R  =+625 

When        x=  —  co         +  +2  variations. 

When         #=-fco          +         +         +0  variation. 
Hence  there  are  two  real  roots  and  four  imaginary  roots. 

4.  How  many  real  roots  has  the  equation  x9  —  15#-|-21=0? 

Arts.    3. 

5.  How  many  real  roots  are  contained  in  the  equation 


6.  How  many  real  roots  are  contained  in  the  equation 

=0?       dm.  2. 


7.  Find  the  number  and  situation  of  the  roots  of  the  equation 

=0. 


X= 

X'=3^2+22a?—  -102 
R=122o?—  393 
R=+ 

Putting    x=  —  co  H-         —        +3  variations. 

ar=-|-co         -t-         +         +         +  0  variation. 
Hence  all  the  roots  are  real. 

To  obtain  the  locality  of  these  roots  there  are  several  principles 
to  guide  us,  there  is  (Art.  180),  but  the  real  limits  are  much 
narrower  than  that  article  would  indicate,  unless  all  the  coeffi- 
cients after  the  first  are  minus,  and  equal  to  the  greatest. 

A  practised  eye  will  decide  nearly  the  value  of  a  positive  root 
by  inspection  ;  but  by  (Art.  183.)  we  learn  that  the  root  of  R, 
or  122#  —  393=0,  must  give  a  value  to  x  intermediate  between 
the  roots  of  the  primitive  equation. 

From  this  we  should  conclude  at  once  that  there  must  be  a 
root  between  3  and  4. 


NEWTON'S  METHOD  OF  APPROXIMATION.  317 

Substituting  3  for  #,  in  the  above  functions,  we  have 

+     —    —     4-2  variations. 
#=4         -h     +     4-     -{-     0  variation. 
Hence  there  are  two  roots  between  3  and  4. 

As  the  sum  of  the  roots  must  be  —  11,  and  the  two  positive 
roots  are  more  than  6,  there  must  be  a  root  near  —  17. 

As  there  are  two  roots  between  3  and  4,  we  will  transform 
the  equation,  (Art.  175),  into  another,  whose  roots  shall  be  less 
by  3;  or  put  a?=3-J-y.     Then  we  shall  have 
X= 
X'= 
R—  I22y  —  27 

R'=+ 

The  value  of  y,  in  this  transformed  equation,  must  be  near 
the  value  of  y  in  the  equation  122y=27,  (Art.  183.)  ;  that  is,  y 
is  between  .2  and  .3 

2/=.2     gives     4~     —    —     +     2  variations. 
7/=.3     gives     4-4~4-4-0  variation. 
Hence  there  are  two  values  of  y  between  .2  and  .3  ;  and,  of 
course,  two  values  of  x  between  3.2  and  3.3. 

We  may  now  transform  this  last  equation  into  another  whose 
roots  shall  be  .2  less,  and  further  approximate  to  the  true  values 
of  xr  in  the  original  equation. 

Having  thus  explained  the  foregoing  principles,  and,  in  our 
view,  been  sufficiently  elaborate  in  theory,  we  shall  now  apply  it 
to  the  solution  of  equations,  commencing  with 

NEWTON'S  METHOD  OF  APPROXIMATION. 

(Art.  189.)  We  have  seenf  m  (Art.  175.),  that  if  we  have  any 
equation  involving  x,  and  put  x=a-{-y,  and  with  this  value 
transform  the  equation  into  another  involving  y,  the  equation 
will  be 


If  a  is  the  real  value  of  x.  then  i/=0,  and  X=0. 


318 


ELEMENTS  OF  ALGEBRA. 


If  a  is  a  very  near  value  to  a?,  and  consequently  y  very 
small,  the  terms  containing  y2,  i/3,  and  all  the  higher  powers  of 
T/,  become  very  small,  and  may  be  neglected  in  finding  the  op- 
proximate  value  of  y. 

Neglecting  these  terms,  we  have 

X-f-X'2/=0, 
Or     .....    y=~  .........  (1) 

In  the  equation  x=a-{-y,  if  a  is  less  than  x,  y  must  be  posi- 
tive ;  and  if  y  is  positive  in  the  last  equation,  X  and  X'  must 
have  opposite  signs,  corresponding  to  (Art.  184.). 

Following  formula  (1),  we  have  an  approximate  value  of  y  ; 
and,  of  course,  of  x.  The  value  of  x,  thus  corrected,  again  call 
«,  and  find  a  correction  as  before  ;  and  thus  approximate  to  any 
required  degree  of  exactness. 

EXAMPLES. 

1.  Given  3a?5+4a?3  —  5x  —  140=0,  to  find  one  of  the  approxi 
mate  values  of  x. 

By  trial  we  find  that  x  must  be  a  little  more  than  2. 
Therefore,  put  x=2+y. 

X  =  3(2)5-{-  4(2)3—  5(2)—  140.  .  .  or  .  .  .  X  =—  22 
X'=15(2)4+12(2)2—  5  ........  or.  .  .  X'=  283. 

X        22 

Whence  y=  —  _=—  =0.07  nearly. 

For  a  second  operation,  we  have 


X=  3(2.07)5-{-  4(2.07)3—  5(2.07)—  140...By  log.  X  =—  -0.854 
X'=15(2.07)4+12(2.07)2—  5    .......  By  log.  X'=  321.82 

Q   &A 

Hence  the  second  value,  or  2/=xHT^-r=0.00265-}- 

o21o.2 

And  ...............    #=2.07265+ 

2.  Given  x*-}-2x2  —  23#=70,  to  find   an  approximate  value 
of  x.  Am.     5.1  3454-. 


HORNER'S  METHOD  OF  APPROXIMATION.  319 

3.  Given    x4 — '3xz-\-75x=lQQQQ,    to    find    an    approximate 
value  of  x.  Ans.     9.886+ 

4.  Given  3jJ* — 35#3 — llx2 — 14rr-f30=0,  to  find  an  approxi- 
mate value  of  x.  dns.       11.998-}-. 

5.  Given  5x3 — 3x* — 2x=1560,  to  find  an  approximate  value 
of  x  rfns.     7.00867+. 

CHAPTER  VI. 
HORNER'S  METHOD  OF  APPROXIMATION. 

(Art.  190.)  In  the  year  1819,  Mr.  W.  G.  Homer,  of  Bath, 
England,  published  to  the  world  the  most  elegant  and  concise 
method  of  approximating  to  roots  of  any  then  known. 

The  parallel  between  Newton's  and  Homer's  method,  is 
this  ;  both  methods  commence  by  finding,  by  trial,  a  near  value 
to  a  root. 

In  using  Homer's  method,  care  must  be  taken  that  the  number, 
found  by  trial,  be  less  than  the  real  root.  Following  Newton's 
method  we  need  not  be  particular  in  this  respect. 

In  both  methods  we  transform  the  original  equation  involving 
#,  into  another  involving  y,  by  putting  x—r-{-y,  as  in  (Art.  175), 
r  being  a  rough  approximate  value  of  #,  found  by  trial. 

The  transformed  equation  enables  us  to  find  an  approximate 
value  of  y,  (Art.  189.). 

Newton's  method  puts,  this  approximate  value  of  y  to  r,  and 
uses  their  algebraic  sum  as  r  was  used  in  the  first  place  ;  again 
and  again  transforming  the  same  equation,  after  each  successive 
correction  of  r. 

Homer's  method  transforms  the  transformed  equation  into 
another  whose  roots  are  less  by  the  approximate  value  of  y ;  and 
again  transforms  that  equation  into  another  whose  roots  are  less, 
and  so  on,  as  far  as  desired. 

By  continuing  similar  notation  through  the  several  transforma- 
tions we  may  have 


320  ELEMENTS  OF  ALGEBRA. 

x=r+y 

y=s-\-z 
z  =  t+z' 
z'=u+z" 
&c.     &c. 

Hence  x=r-\-s-\-t,  &c. ;  r,  s,  t,  &c.,  being  successive  figures 
of  the  root.  Thus  if  a  root  be  325,  r=300,  s=20,  and  /=5. 

On  the  principle  of  successive  transformations  is  founded  the 
following 

RULE  for  approximating  to  the.  true  value  of  a  real  root 
of  an  equation. 

1st.  find  by  Sturm1  s  Theorem,  or  otherwise,  the  value  of 
the  first  one  or  two  figures  of  the  root,  which  designate  by  r. 

2d.  Transform  the  equation  (Art.  175),  into  another  whose 
roots  shall  be  less  by  r. 

3d.  With  the  absolute  term  of  this  transformed  equation  for  a 
dividend,  and  the  coefficient  of  y  for  a  divisor,  find  the  next 
figure  of  the  root. 

4th.  Transform  the  last  equation  into  another,  whose  roots 
shall  be  less,  by  the  value  of  the  last  figure  determined;  and 
so  proceed  until  the  whole  root  is  determined,  or  sufficiently 
approximated  to,  if  incommensurable. 

NOTE  1.  In  any  transformed  equation,  X  is  a  general  symbol 
to  represent  the  absolute  term,  and  X'  represents  the  coefficient 
of  the  first  power  of  the  unknown  quantity.  If  X  and  X'  be- 
come of  the  same  sign,  the  last  root  figure  is  not  the  true  one, 
and  must  be  diminished. 

NOTE  2.  To  find  negative  roots,  change  the  sign  of  every  alter- 
nate term,  (Art.  178.) :  find  the  positive  roots  of  that  equation, 
and  change  their  signs. 

(Art.  191.)  We  shall  apply  this  principle,  at  first,  to  the  solu- 
tion of  equations  of  the  second  degree ;  and  for  such  equations 
as  have  large  coefficients  and  incommensurable  roots,  it  will  fur- 
nish by  far  the  best  practical  rule. 


HORNER'S  METHOD  OF  APPROXIMATION.  321 

EXAMPLES. 
1.  Find  an  approximate  root  of  the  equation 

x*-\-x— 60=0. 

We  readily  perceive  that  x  must  be  more  than  7,  and  less 
than  8,  therefore  r=7. 

Now  transform  this  equation  into  another  whose  roots  shall  be 
less  by  7. 

Operate  as  in  (Art.  175.),  synthetic  division 

i  i  —.60     (7 

7  56 

8  —  4 
7 

15 


Trans.,  eq.       y*  -f-  15y  —  4=0 

Here  we  find  that  y  cannot  be  far  from  T4^,  or  between  .2  and 
.3 ;    therefore  transform  the  last  equation  into  another  whose 

roots  shall  be  .2  less  ;  thus, 

s 

1  15  —4         (.2 

0.2  3.04 

15.2  —0.96 

2 

15.4 

The  second  transformed  equation,  therefore,  is 
z2-f- 15.4*— 0.96=0 

.96 

To  obtain  an  approximate  value  of  z,  we  have  ^— —  or  0.06. 

15.4 

In  being  thus  formal,  we  spread  the  work  over  too  large  a 
space,  and  must  inevitably  become  tedious.  To  avoid  these  diffi- 
culties, we  must  make  a  few  practical  modifications. 

1st.  We  will  consider  the  absolute  term  as  constituting  the 
second  member  of  the  equation ;  and,  in  place  of  taking  the 
algebraic  sum  of  it,  and  the  number  placed  under  it,  we  will  take 
their  difference. 


322  ELEMENTS  OF  ALGEBRA. 

2d.  We  will  not  write  out  the  transformed  equations  ;  that  is, 
not  attach  the  letters  to  the  coefficients ;  we  can  then  unite  the 
whole  in  one  operation. 

3d.  Consider  the  root  a  quotient ;  the  absolute  term  a  dividend, 
and,  corresponding  with  these  terms,  we  must  have  divisors. 

In  the  example  under  consideration,  8  is  the  first  divisor ;  15 
.s  thejirst  trial  divisor ;  15.2  is  the  second  divisor,  and  15.4  is 
the  second  trial  divisor ;  15.46  is  the  third  divisor,  &c. 

Let  us  now  generalize  the  operation.  The  equation  may  be 
represented  by 


Transform  this  into  another  whose  roots  shall  be  less  by  r  ;  that 
equation  into  another  whose  roots  shall  be  less  by  s,  &c.,  &c. 

SYNTHETIC    DIVISION. 

1  a  n  (  r-j-s 

r  ar-f-r8 


1st  divisor,     ....  a-\-r  n' 

r  (a-\-2r-}-s}s 

1st  trial  divisor,  .  .  a-f-2r  n" 

0-t-s  &c. 


2d  divisor,     .  .  .  .a-\-2r-}-s 

s 


2d  trial  divisor,  .  .a+2r+2« 
&c. 

In  the  above  we  have  represented  the  difference  between  n 
and  (ar-{-»'2)  by  n',  &c.  As  n',  n",  n'",  &c,,  with  their  corre- 
sponding trial  divisors,  will  give  5,  t,  u,  &c.,  the  following  for- 
mula  will  represent  the  complete  divisors  for  the  solution  of  all 
equations  in  the  form  of 


IIORJNER'S  METHOD  OF  APPROXIMATION  3^3 

1st  divisor,   ....     ±a+  r 

add    ....  r-\-  s 


2d  divisor, 
add 


3d  divisor,   ....     ±a+2r+2*-f-  t 

add    ....  t-\-u 


4th  divisor,  .... 

&c.  &c.  &c. 

Equations  which  have  expressed  coefficients  of  the  highest 
power,  as 


the  formulas  will  be  : 

1st  divisor,    ....     ±a-h  cr 

add  ....  cr-f-  cs 


2d  divisor,    ....     ±a-f  2cr+  cs 

add   ....  cs-\-ct 


3d  divisor,   ....     ±a-f  2cr-j-2cs-t-c* 
&c.  &c. 

To  obtain  trial  divisors  we  would  add  cr  only,  in  place  of 
(cr+cs),  &c. 

We  will  now  resume  our  equation  for  a  more  concise  solution. 


1 

7 

n               r 
60           (  7. 
56 

stu 
262 

1st  divisor,  .  . 
add  .  . 

8 
7.2 

4 

304 

2d  divisor,  .  . 
add  .  . 

15.2 
26 

96 
9276 

3d  divisor,  .  . 
add  .  . 

15.46 

62 

324 

31044 

4th  divisor,  .  .     15.522  1356 


324  ELEMENTS  OF  ALGEBRA. 

We  can  now  divide  as  in  simple  division,  and  annex  the  quo- 
tient figures  to  the  root,  thus : 

15522)1356     (08734 
124176 

11424 
10865 


559 

465 

94 
Hence #=7.2620873-f- 

2.  Find  #,  from  the  equation  xz — 700#=59829. 
On  trial,  we  find  x  cannot  exceed  800 ;  therefore,  r— 700. 

n       rst 

-h  r  -700+700=     0  59829(777 

r+  s  700-H  70=770  00)000 


=770  770)59829=n' 

77  5390 

=847  847)5929=n" 

5929 


Hence, #=777. 

3    Find  #  from  the  equation  x2 — 1283#= 16848. 

By  trial,  we  find  that  #  must  be  more  than  1000,  and 

than  2000;  therefore,  r=1000. 

n      rstu 

«4-  r=      —283  16848(1296 

H-  5=    1200  —283 

a-r-2r4-  ~s=    917          917)2998 
s+t    290  1834 


=  1207         1207)11644 
96  10863 


1303         1303) 
Hence, 


HORNER'S  METHOD  OF  APPROXIMATION.  305 

4.  Given  x2—  5;r=8366,  to  find  x. 

By  trial,  we  find  x  must  be  more  than  90,  and  less  than  100 
Therefore,        a-f-  r      =   85  )  8366  (  Q4=x 


r+s  .  .  94 
«+2H-s=179 


765 

716 
716 


5.  Find  a?,  from  the  equation  x2  —  375#+  1904=0. 
Here  the  first  figure  of  the  root  is  5. 


5 
—375 

1st  divisor,  — 370. 
5.1 

2d  divisor,  — 364.9 
.14 


—1904(5.1480052207 
—1850 

—5400 
—3649 


3d  divisor,  —364.76 
48 


4th  divisor,  —364.712 
8 


5th  divisor,  — 364.7039 


—175100 
—145904 

— 2919600 
—2917696 

—190400 
—  1823519 

—80491 
—72941 

—7550 
—7294 


—256 
—255 

— 1 

6.  Given  ^+7^—1194=0,  to  find  x.      Ans.  31.2311099. 

7.  Given  x3— 21^=214591760730,  to  find  x.     rfns.  463251. 

It  might  be  difficult  for  the  pupil  to  decide  the  value  of  r,  as 
applied  to  the  last  example,  without  a  word  of  explanation :  x 
must  be  more  than  the  square  root  of  the  absolute  term,  that  is, 


326  ELEMENTS  OF  ALGEBRA. 

more  than  400000  ;  then  try  500000,  which  will  be  found  too 
great. 

(Art.  192.)  When  the  coefficient  of  the  highest  power  is  not 
unity,  we  may  (if  we  prefer  it  to  using  the  last  formulas  for  di- 
visors), transform  the  equation  into  another,  (Art.  166.),  which 
shall  have  unity  for  the  coefficient  of  the  first  term,  and  all  the 
other  coefficients  whole  numbers. 


8.  Given  7x*—  3#=375. 

Put  #=-    and  we  shall  have  y2  —  3^=262  5. 

One  root  of  this  equation  is  found  to  be  52.7567068+,  one- 
seventh  of  which  is  7.536672+  ;  the  approximate  root  of  the 
original  equation. 

9.  Given  7#2  —  83#+  187=0,   to  find  one  value  of  a?. 

Jlns.     3.024492664 

1O.  Given  x*  —  T3Ta?=8,  to  find  one  value  of  x. 

Am.     2.96807600231 


11.  Given  4#2+J#=^,  to  find  one  value  of  x. 

Jlns.    Jlns.     .14660+ 

12.  Given    \x*-\-\x—  T7T,  to  find  one  value  of  x. 

Jlns.     .6042334631 

13.  Given  115  —  3x*  —  7#=0,  to  find  one  value  of  x. 

Jlns.     5.13368606 

(Art.  193.)  We  now  apply  the  same  principle  of  transforma- 
tion to  the  solution  of  equations  of  the  third  degree. 

EXAMPLES. 
1.  Find  one  root  of  the  equation 


We  find,  by  trial,  that  one  root  must  be  between  3  and  4. 


HORNER'S  METHOD  OF  APPROXIMATION. 


327 


1st  Transformation. 

1       —1 
3 

3 

~5 
3 

~S 

70 

*76 
15  J 

91  =X' 

r 
—300        (3. 

228 

—  72=X 

....                   _-72_A 

to 

o. 

H 
1 

I           8 
0.7 

91  ' 
6.09-| 

s 
—  72        (0.7 
67.963 

8.7 

7 

*97.09  , 
6.58J 

—     4.037=X 

9.4 

7 

To.7 

103.67=X' 

1         10.1 
.03 

103.67 
.3039-} 

t 
—    4.037     (0.03 
—     3.119217 

10.13 
3 

*103.9739 
.3048J 

—     0.917783=X 

10.16 
3 

HU9" 

104.2787=X' 

1         10.19 
.008 

104.2787 
.081584 

u 
0.917783(0.008 
0.834882272 

10.198 
8 

10.206 

8 

10.214 

*104.360284 
81648 

104.441932 

.082900728 

328  ELEMENTS  OF  ALGEBRA. 

The  terms  here  marked  X   are  trial  divisors ;  we  have  pro 
fixed  stars  to  the  numbers  that  we  may  call  complete  divisors. 
We  rest  here  with  the  equation 

(z")3+10.214(z//)24-104.4419z"— 0.0829=0. 

The  value  of  z"  is  so  small  that  we  may  neglect  all  its  powers, 

except  the  first,  and  obtain  several  figures  by  division,  thus : 

104  )  829  (  797 

728 


74 

r  s tu 

Hence,  one  approximate  value  of  #  is     ....     3.738797+. 
(Art.  193.)  We  may  make  the  same  remarks  here  as  in  (Art. 
191.),  and,  as  in  that  article,  generalize  the  operation. 

Let  xs -\-J2z? -\-Bx-N  represent  any  equation  of  the  third 
degree,  and  transform  it  into  another  whose  roots  shall  be  r  less ; 
thus, 

1  A  B  =    N  (  r 

r 


N' 


The  transformed  equation  is 


If  we  put        ( 
and 

we  shall  have    .  .  .     y*-\-*ft'yz-{-B'y=Nt,    an  equation  similar 
to  the  primitive  equation. 

If  we  transform  this  equation  into  another  whose  roots  shall  be 
less  by  s,  we  shall  have 


Or,  .  .   z*-\-£"z*-\-B"z=N"  ;    an  equation  also  similar  to 


HORNER'S  METHOD  OF  APPROXIMATION.  329 

the  first  equation.  And  thus  we  may  go  on  forming  equation 
after  equation,  similar  to  the  first,  whose  roots  are  less  and  less. 

The  quantities  N',  N",  &c.,  are  the  same  as  X  in  our  previous 
notation,  and  the  quantities  B',  B",  &c.,  are  the  same  as  the 
general  symbol,  X' ;  but  we  have  adopted  this  last  method  of 
notation  to  preserve  similarity. 

Observe,  that  as  (r-{-JT)r-{-JB  is  the  first  complete  divisor,  N 
is  the  number  considered  as  a  dividend  and  r  the  quotient ;  and 

therefore...  r^-^—,  nearly. 

The  next  equation  gives  us 

'  ' 

('  " 


JV»  N"          very  nearly. 

And  the  next,  /= 


N'"  N" 


&c.  =         &c.  &c. 

The  denominators  of  these  fractions  are  considered  complete 
divisors,  and  the  quantities,  B',  B",  B'",  are  considered  trial 
divisors.  The  further  we  advance  in  the  operation  the  nearer 
will  the  trial  and  true  divisors  agree. 

Before  the  operation  is  considered  as  commenced,  we  must 
find  the  first  figure  of  the  root  (r)  by  trial.  Then  the  operator 
can  experience  no  serious  difficulty,  provided  he  has  in  his  mind 
a  clear  and  distinct  method  of  forming  the  divisors ;  and  these 
may  be  found  by  the  following 

RULE.  1st.  Write  the  number  represented  by B,  and  directly 
under  it,  write  the  value  of  r(r+A) ;  the  algebraic  sum  of  these 
two  numbers  is  the  first  complete  divisor. 

2d.  Directly  under  the  first  divisor  write  the  value  of  r2,  and 
the  sum  of  the  last  three  numbers  is  B1,  or  the  first  trial  divisor. 

3d.  Find  by  trial,  as  in  simple  division,  how  often  B'  is  con- 
tained in  N,  calling  the  first  figure  s,  (making  some  allowance 
for  the  augmentation  o/'B'),  and  s  will  be  a  portion  of  the 
root  under  trial. 
28 


330  ELEMENTS  OF  ALGEBRA. 

4th.  Take  the  value  of  the  expression  (3r+s-[-A)«  and  add 
it  to  the  first  trial  divisor  ;  the  sum  is  the  second  divisor  (if 
we  have  really  the  true  value  of  s). 

IN    GENERAL   TERMS; 

Under  any  complete  divisor,  write  the  square  of  the  last  figure  of 
the  root  ;  add  together  the  three  last  columns,  and  their  sum  is 
the  next  trial  divisor.  With  this  trial  divisor  decide  the  next 
figure  of  the  root. 

Take  the  algebraic  sum,  of  three  times  the  root  previously 
found,  the  present  figure  under  trial,  and  the  coefficient  Jl,  and 
multiply  this  sum  by  the  figure  under  trial,  and  this  product, 
added  to  the  last  trial  divisor,  gives  the  next  complete  divisor. 

EXAMPLES. 

1.  Given  a?3+2a^+3a?=  13089030,  to  find  one  value  of  x. 
By  trial,  we  soon  find  that  x  must  be  more  than  200  and  less 
than  300  ;  therefore  we  have 

r=200,     .#=2,    .5=3. 
By  the  rule, 

N       rs 
B  .......     3    40403  )  13089030  (  235 

40400  -          80806 


-j 


1st  divisor  .......  40403  f   139763  )  500843 

r2  ........  40000J         419289 


1st  trial  divisor  .  .  B'  =120803         163108  )    815540=^ 
(3H-s-M)s    •  •  •    18960^  815540 

2d  divisor  .......  139763  f 

s2  ........        900J 

2d  trial  divisor  •  •  JB"=I  59623 
•      3485 


3d  divisor  .......  163108 

Hence,  ........  .  ...............  z=235, 


HORNER'S  METHOD  OF  APPROXIMATION.  331 

3.  Given  ^-(-173^=14760638046,  to  find  one  value  of  x 
Here  .#=0,  I?=173,  and  we  find,  by  trial,  that  x  must  be 
more  than  2000,  and  less  than  3000  ;  therefore  r=2000. 
B  ........  173 

4000000 


1st  divisor  .........    4000173  f 

r2    .........    4000000J 


B'  ........  12000173 

2560000 


2d  divisor  .........  14560173  f 

s2  .........    leooooJ 


B"    .......  17280173 

.      362500> 


3d  divisor  .........  17642673  f 

2500  J 


B'"  .......  18007673 

....        22059 


4th  divisor    ........  18029732 

4000173  )  14760638046  (  2453=z 
8000346 


14560173  )     67602920     =N' 
58240692 


17642673  )      93622284  =N" 
88213365 


18029732  )        54089196=^'" 
54089196 


3.  Given  a?3+2a^  —  23#=70,   to  find  an  approximate  value 
of  x.  rfns.    #=5.134578-K 

4.  Given  x*  —  17a^-i-42iC=185,  to  find  an  approximate  value 
of  x.  Jns.     r==  15.02407. 


332  ELEMENTS  OF  ALGEBRA. 

(Art.  194.)  When  the  coefficient  of  the  highest  power  is  not 
unity,  we  may  transform  the  equation  into  another,  (Art.  166.), 
in  which  the  coefficient  of  the  first  term  is  unity,  and  all  the 
other  coefficients  whole  numbers  ;  but  it  is  more  direct  and  con- 
cise to  modify  the  rule  to  suit  the  case. 

If  the  coefficient  of  the  first  power  is  c,  the  first  divisor  will 
be  (cr-f^)r-f-^,  in  place  of  (r+A)r-\-B. 

In  place  of  (3r-{-s-\-JI)s,  to  correct  the  first  trial  divisor,  we 
must  have  (3cr-\-cs-\-tf)s  ;  and,  in  genera],  in  place  of  using  3 
times  the  root  already  found,  we  must  use  3c  times  the  root  ;  and, 
in  place  of  the  square  of  any  figure,  as  r2,  s2,  &c.,  we  must  use 
cr2,  cs2,  &c. 

EXAMPLES. 

5.  Find  one  root  of  the  equation,  3:e3-{-2;r2-}-4;r=75, 

By  trial,  we  find  that  x  must  be  more  than  2,  and  less  than  3; 

therefore  ,    0  a     0  »     A 

r=2,         c=3,         •#=2,         /j—4. 

N      r  stu 
B  ........    4.  75  (  2.577 


16.  > 


1st  divisor 20. 

cr2 12.  J 


40 

35=7V' 
29375 


B1    .......  48.  5625=^" 

.  .10.75  5038579 


2d  divisor  ......  58.75  f  .  586421  =N'" 


cs 


75J  .517301099 


B"  .......  70.25  69119901 


3d  divisor      .....  71.9797  [   Continue,  by  simple  division,  thus 

ct*  .......  _  Jfy  739  )  6911  (  935 

B>"  .......  73.7241  6651 

•        176057  ~ 


4th  divisor     .....  73.900157  221 

Hence,    .................  z  =2.  577935  -f. 

*   R  is  a  symbol  to  represent  the  entire  root,  as  far  as  determined. 


HORNER'S  METHOD  OF  APPROXIMATION.  333 

6.  Find  one  root  of  the  equation,     5#3  —  Qx?-\-3x—  —  85. 

rfns.     x=  —  2.16399-—. 

7.  Find  one  root  of  the  equation,    \2x*-\-x*  —  5a?=330. 

dns.     #=3.0364754- 

8    Find  one  root  of  the  equation,  5#3-f  9#2  —  7#=2200. 

Ans.     #=7.107353G-f-. 

9.  Find  one  root  of  the  equation,   5#3—  3a?2  —  2.r=1560. 

Ans.     #=7.0086719-1-  . 

(Art.  195.)  This  principle  of  resolving  cubic  equations  ma)' 
be  applied  to  the  extraction  of  the  cube  root  of  numbers,  and 
indeed  gives  one  of  the  best  practical  rules  now  known. 

For  instance,  we  may  require  the  cube  root  of  100.  This 
gives  rise  to  the  equation 


in  which  .#=0,  and  5=0,  and  the  value  of  x  is  the  root 
sought. 

As  Ji  and  B  are  each  equal  to  zero,  the  rule  under  (Art.  193.) 
may  be  thus  modified. 

1st.  Keeping  the  symbols  as  in  (Art.  193.),  and  finding  r  by 
trial,  r2  will  be  the  first  divisor,  and  3r2  is  B',  or  the  first  TRIAL 
divisor. 

2d.  By  means  of  the  dividend  (so  called),  and  the  first  trial 
divisor,  we  decide  s  the  next  figure  of  the  root. 

3d.  Then  (3r-{-s)s  ;  that  is,  three  times  the  portion  of  the 
root  already  found,  with  the  figure  under  trial  annexed,  and 
the  sum  multiplied  by  the  figure  under  trial,  will  give  a  sum, 
which,  if  written  two  places  to  the  right,  under  the  last  trial 
divisor,  and  added,  will  give  the  next  complete  divisor. 

4th.  After  we  have  made  use  of  any  complete  divisor,  write 
the  square  of  the  last  quotient  figure  under  it  ;  the  sum  of  the 
three  preceding  columns  is  the  next  trial  divisor  ;  which  use, 
and  render  complete,  as  above  directed,  and  so  continue  as  far  as 
necessary.* 

*  In  case  of  approximate  roots  after  three  or  four  divisors  are  found,  we 
may  find  two  or  three  more  figures  of  the  root,  with  accuracy,  by  simple  division. 


334  ELEMENTS  OF  ALGEBRA. 

We  may  now  resolve  the  equation 


1st  divisor  •  • 

(3r-f-s>    . 
2d  divisor  •  • 


.  •  16 

.  .48 
.  .    756 


.  .  5556 


r  stu 

100  (  4.6415889-t- 
64 

36 
33336 


B" 6348 

(3R+t)t    .  •  •      5536] 

3d  divisor  ....  640336  f 

? 16  J 

B"' 645888 

«.  •  .        13921 


4th  divisor.  •  •  .64602721 


64616643 
646166 


2.  Extract  the  cube  root  of 
1st  divisor  *  *    64 
B'  •  •  •  192 


(3r-f-s)s    •    1729 


2d  divisor 


B"  . 


2664000 
2561344 

102656 
64602721 

38053279 
32308321 

5744958 
5169331 

575627 
516933 

58694 
58154 

673373097125. 

N  rstu 
673373097125  (8765 
512 


3d  divisor 


B'" 


20929  \ 
49  j 

161373 

146503                  rNOTE.-To  deter- 

22707 
15696] 

mine  s,  we  have 

14870097        1     i92)1613( 

13718376          |  Some    allowance 

2286396  \ 
36J 

11  51721  1251    crease  of  192. 

1151721125 

2302128 
131425 

230344225 


HORNER'S  METHOD  OF  APPROXIMATION.  335 

3.  Extract  the  cube  root  of  1352605460594688. 

tins.     110592. 

4.  Extract  the  cube  root  of  5382674.      tins.  175.25322796. 

5    Extract  the  cube  root  of  15926.972504. 

Ans.     25.16002549. 

6.  Extract  the  cube  root  of  91632508641. 

Ans.     4508.33859058. 

7.  Extract  the  cube  root  of  483249.         Am.     78.4736142 

(Art.  196.)  The  method  of  transforming  an  equation  into  an- 
other, whose  roots  shall  be  less  by  a  given  quantity,  will  resolve 
equations  of  any  degree  ;  and  for  all  equations  of  higher  degrees 
than  the  third,  we  had  better  use  the  original  operation,  as  in 
(Art.  192.),  and  attempt  no  other  modification  than  conceiving  the 
absolute  term  to  constitute  the  second  member  of  the  equation ; 
and  the  difference  of  the  numbers  taken  in  the  last  column  in 
place  of  their  algebraic  sum. 

The  following  operation  will  sufficiently  explain : 
1.  Find  one  value  of  x  from  the  equation 


r 

1   ±0 

—3 

75 

=  10000    (9 

9 

81 

702 

6993 

9 

78 

777 

3007=^V 

9 

162 

2160 

18 

240 

2937 

9 

243 

27 

483 

9 

36 

(Continued  on  the  next  page.) 


336 


ELEMENTS  OF  ALGEBRA. 


1       36 

483. 

2937. 

9 

=3007.    (    0.8 

to 

c- 

t   -^ 

.8 

29.44 

409.952 

2677.5616 

36.8 

512.44 

3346.952 

329.4384=^" 

ansformation. 

8 

3776 

8 

38.4 

30.08 

542.52 
30.72 

573.24 

434.016 

coefficients  to  their  nearest 

3780.968 
Take  the 

8 

unit. 

39.2 

t 

1       39 

573 

3781 

=  329.4384  (  0.08 

+ 

3 

46 

306.16 

39 

576 

3827 

23.2784=^'" 

+ 

3 

46 

39 

579 

3873 

u 

1       39 

579 

3873 

=     23.2784(0.00600-1- 

3 

23.256 

3876 

224 

3 

3879 
Hence, a-=9.88600+. 

N.  B.  We  went  through  the  first  and  second  transformations  in  full.     Had 

we  been  exact,  in  the  third,  we  should  have  added  .08  to  39.2,  and  multiplied 

their  sum,  (39.28),  by  .08,  giving  3.1424 ;  we  reserve  3.  only  to  add  to  the 

next  column.     By  a  similar  operation  we  obtain  46.  to  add  to  the  next  column. 

EXAMPLES. 

1.  Given  x— &— x3— x4-J-500=0,  to  find  one  value  of  a;.   Ans.  4.46041671 

2.  Given  a;4—  5.r3-}-9;r=2.8,  to  find  one  value  of  x.     Ans.  .32971055072 

3.  Given  20#-{-ll;r2-j-9z3— ^=4,  to  find  one  value  ofx. 

Ans.  .17968402502 

4.  Required   the  5th  root  of  5000;  or,  in  other  terms,  find  one  root  of  the 
equation  ^=5000.  Ans.     5.49280-}- 


5.  Given  2?= 


to  find  one  value  of  x. 


Ans.    2.120003355 


THEORY  OF  EQUATIONS-SUM  OF  COEFFICIENTS.     337 

In  following  out  general  principles, we  have,  thus  far,  omit- 
ted some  artifices  which  apply  in  particular  cases,  or  depend 
on  particular  circumstances,  which  we  will  now  set  forth,  as 
they  will  enlarge  the  views  of  all  who  pay  attention  to  them. 

The  first  circumstance  to  be  observed  is  the  sum  of  the 
numerical  coefficients. 


(Art.  197.)  Transpose  every  term  to  the  first  member  of  the 
equation,  and  if  the  sum  of  the  coefficients  is  zero,  we  are  sure 
that  unity  is  one  of  the  roots  of  the  equation. 

For  example,  we  are  sure  that  the  equation 
a;4+7a;3—  1332—-79*+84==0, 

has  unity  for  one  of  its  roots,  because  l-J-7  —  13  —  79+84=0. 
Assume  x=l,  and  substitute  1  for  x  in  the  equation,  and  the 
equation  will  be  verified.  But  assume  x,  a  little  greater  or  a 
little  less  than  1,  and  substitute,  and  the  equation  cannot  be 
verified  ;  therefore  the  following  principle  is  established  : 

In  any  equation  having  numerical  coefficients,  if  the  sum  of  the 
coefficients  is  zero,  -\-l  is  one  of  the  roots  of  the  equation. 

We  can  now  divide  the  given  equation  by  (x  —  1)  and  thus 
depress  it  one  degree. 

For  another  example  we  will  recall  the  4th,  on  page  266, 
which  is 


Here  1  —  13-(-12=0,  whence  x=\.  The  answer  on  page 
266  is  3  ;  that  is  another  root.  The  third  root  must,  therefore, 
be  —  4  ;  the  sum  of  all  the  roots  must  be  0,  because  the  second 
term  of  the  equation  is  zero.  (Art.  157,  second  observation.) 

We  perceive  at  once,  that  unity  is  one  root  in  each  of  the 
following  equations  : 


2.  x3—  Qx^  +  Ux—  6=0. 

,  3.  x*—Qx3Jr\4x2±8x—}5—Q. 

4.  X3_  i3^2_j_39a._  27=0. 

(Art.  198.)     When   the   absolute   term  of  an   equation    is 
numerically  too  large  (whether  it  be  plus  or  minus),  to  make 
29 


338  ELEMENTS  OF  ALGEBRA. 

the  sum  of  the  coefficients  zero,  we  can  diminish  it,  as  well  as 
other  coefficients,  by  the  following  transformation. 
For  instance  :  In  the  equation 


the  sum  of  the  coefficient  is  obviously  not  equal  to  zero  ;  70  is 
too  large,  but  we  can  diminish  it  by  placing  x=2y,  or  x=ny, 
and  substituting  this  value  of  x  in  the  equation,  we  have 

H3y3  —  iOw2 
Dividing  by  n3  we  obtain 


n          n 

Now,  we  can  assume  n  equal  to  2,  3,  4,  or  any  other  number 
whatever. 

Let  w=2,  and  the  preceding  equation  becomes 

y'-5ys+"H-™=o.  (i) 

let»=-2, 


Let  7i=5,  and     y3-^+y  +=0.  (3) 

Zo        ~o 

The  sum  of  the  coefficients  in  each  of  the  equations  (2)  and 
(3),  is  zero.     Therefore,  y  equals  1  in  each  of  them. 

But  x=ny,  therefore  z—  —  2,  and  #=5, 

which  are  two  of  the  three  roots  in  the  original  equation. 

But  the  sum  of  the  three  roots  must  be  10,  by  the  genera] 
theory  of  equations  ;  therefore,,  the  third  root  must  be  7, 

Because,  —  2+5-f  7=  1  0. 

If  we  take  the  first  equation  in  (Art.  197),  and  depress  it 
one  degree,  by  dividing  by  (x  —  1),  we  shall  obtain  the  equation 

z*-±-8x2—ox—U=Q.  (a) 

Here  it  is  obvious  that  the  coefficients  require  depressing  to 
make  their  sum  zero. 

Therefore,  place  x=nP,  and 

p3  i  8jp2_l  p  __84  =a  ^ 

w          n2         n3  v  ' 


THEORY  OF  EQUATIONS—  SUM  OF  COEFFICIENTS.     339 

Making  n=3,  this  equation  becomes 

2_5p  _28=0 
99 

Here  ?+?i—  -—  —  =0.     Whence  P=l,  but  »=3, 
99      9      9 

and  z=nP=3,  another  of  the  roots  of  the  original  equation. 

We  may  now  depress  equation  (a)  one  degree,  by  dividing 
by  (x  —  3),  and  then  find  the  other  two  roots. 

Or,  we  may  assume  n—  —  4,  and  substitute  this  value  for  n 
in  (b),  then  that  equation  becomes 


Here,  because  H~ff—  r56-h?6=°-  ^=1,  and  x=—  4, 
which  is  a  third  root  of  the  original  equation.  x=  —  7  is  the 
fourth  root. 

Thus  we  can  find  all  the  roots  in  the  equation  contained  in 
the  last  article. 

If  it  were  our  object  to  transform  an  equation   so  as  to 

increase  the  coefficients    in    place    of    diminishing    them,     as 

p 
in  this  article,  we  should  place  #=  —  ,  but  this   has  already 

n 

been  shown  in  (Art.  166).  • 

This  method  of  solution  may  be  considered  a  sequence  to 
Newton's  method  of  dividers  as  shown  in  (Art.  147). 

(Art.  199.)  An  equation  which  has  one  for  a  root,  can  bo 
changed  into  another,  which  will  have  minus  one  for  a  root,  by 
changing  its  second,  and  every  alternate  sign.  (Art.  178.) 

Consequently,  then,  an  equation  which  has  minus  one  for 
one  of  its  roots,  can  be  changed  into  another,  which  will  have 
plus  one  for  one  of  its  roots,  by  changing  the  sign  of  its  second, 
and  every  alternate  term. 

Therefore,  if  changing  the  sign  of  the  second,  and  every  alter- 
nate term  in  any  equation  ivill  make  the  sum  of  the  coefficients 
ZERO,  one  root  of  that  equation  must  be  —  1. 

For  example,  the  sum  of  the  coefficients  of  the  equation 
4.c3—  20;r2-{-ll.r+35=0,  is  not  zero; 


340  ELEMENTS  OF  ALGEBRA. 

but  change  the  signs  as  above  directed,  and  we  have  the 
equation, 

4z8-f20#2-)-l  1^—35=0, 

and  the  sum  of  the  coefficients  is  now  zero,  and  consequently 
unity  is  one  root  of  this  equation,  and  therefore  — 1  is  a  root 
of  the  first  mentioned  equation. 

Hence,  in  seeking  to  solve  any  equation  of  this  kind,  if  the 
sum  of  the  coefficients  is  not  zero,  we  may  change  the  sign  of 
the  second,  and  each  alternate  term,  and  if  the  sum  is  then 
zero,  — 1  may  be  taken  as  one  root  of  the  equation  ;  and  then 
we  may  depress  the  equation  one  degree  by  dividing  by  (x-\-\). 


(Art.  200.)  Another  circumstance  to  be  observed  in  rela- 
tion to  coefficients,  is  that  of  their  recurrence,  producing 
what  is  called 

RECURRING  EQUATIONS. 

In  a  recurring  equation  the  coefficients  of  those  terms  which  are 
equally  distant  from  the  extreme  terms,  are  numerically  equal. 
Thus    x5—  3x*+7x3+7x2—  3a?-fl=0,  (1) 


are  all  recurring  equations  of  an  odd  degree. 

If  we  substitute  —  1  for  x,  in  each  of  the  equations  (1)  and 
(2),  they  will  be  verified;  and  -f-  1  put  for  x  will  verify  (3) 
and  (4).  Therefore,  any  equation  of  an  odd  degree  has  either 
—  1  or  _j_  i  for  one  of  its  roots,  and  the  equation  itself  is  redu- 
cible to  an  even  degree,  by  dividing  by  (x-\-\  )  or  by  (  x  —  ]  ). 

Minus  one  is  a  roof,  when  the  like  coefficients  have  the  same  sign; 
and  plus  one  is  a  root  when  the  like  coefficients  have  opposite  signs. 

(Art.  201.)  Recurring  equations,  of  an  even  degree,  in  ivhich 
the  like  coefficients  have  opposite  signs,  and  ivhoss  middle  term  is 
ivanting,  are  divisible  by  (x2  —  1),  and  therefore  +1,  and  —  1, 
are  both  roots  of  every  such  equation. 


THEORY  OF  EQUATIONS—  SUM  OF  COEFFICIENTS.     341 
The  following  is  an  equation  of  this  description  : 


which  can  be  put  in  the  following  form  : 

(x*—l)+8x(x*—l)-\-llx2(x2—l)=Q. 

and  this  is  obviously  divisible  by  (x2  —  1).     Whence  we  may 
place  x2  —  1=0,  and  x=l,  or  —  1. 

And  thus  the  principle  enunciated  in  this  article  is  estab- 
lished. Dividing  the  equation  by  (x2  —  1),  will  reduce  it  two 
degrees  lower,  and  leave  it  an  equation  of  an  even  degree. 

(Art.  202.)  Every  recurring  equation  of  an  even  degree,  above 
the  second,  can  be  reduced  to  an  equation  of  half  that  degree,  by  the 
following  artifice. 

We  take  the  following  equation  for  an  example  : 

#6_7a.5_|_82;4  —  9z3  -far2—  72+1=0.          (1) 

Divide  each  term  by  the  square  root  of  the  highest  power  of 
the  unknown  quantity.  In  this  case  divide  by  x3,  and  we  have 

.  (2) 


Which  can  be  put  in  the  following  form  : 


Now  assume,       /Vfi'Wz.     Then,    fx2-\-^\=z2—  2. 

And  Sx3-\--—  )=z3  —  3z.     Whence  (3)  becomes 

v     as3/ 

(23_32)—7(s2—  2)4-8^—  9=0,  (4) 

an  equation  of  the  third  degree. 

Thus  (1),  an  equation  of  the  6th  degree  is  reduced  to  (4),  an 
equation  of  half  that  degree  ;  and  this  process  will  reduce  any 
other  recurring  equation  of  an  even  degree  to  another  of  half 
that  degree. 

(Art.  203.)  A  recurring  equation  of  the  4th  degree  can  be 
reduced  to  a  quadratic  by  the  following  rule. 

Divide  by  the  coefficient  of  x4  ,  and  transpose  the  term  containing 
-x1.  Add  to  each  member  +2#-.  Use  the  plus  sign  when  the  signs 


342  ELEMENTS  OF  ALGEBRA. 

of  the  second  and  fourth  term  are  alike,  and  use  the  minus  sign 
when  tJtey  are  unlike.  Also  add  to  each  member  the  square  of 
half  the  term  containing  x.  Extract  the  square  root  of  each  mem- 
ber, and  the  result  is  an  equation  of  the  second  degree. 

EXAMPLES. 


a  a 


Add 


.x,  a  quadratic. 

We  are  indebted  to  Professor  Stevens,  of  Greenmount  Col- 
lege, Indiana,  for  this  article.  It  is  original  with  him. 

BINOMIAL  EQUATIONS. 

(Art.  204.)  Binomial  equations  have  but  two  terms,  such  as 

a;—  1=0,     z2—  1=0,     z3—  1=0,  &c. 
Or,  such  as     #±a=0,     #"±an  =0,  &c.,  &c. 

Every  binominal  equation  is  also  a  recurring  equation,  for 
the  coefficient  certainly  recurs. 

A  binomial  equation  has  apparently  but  one  root,  but  a  full 
investigation  will  discover  as  many  roots  as  there  are  units 
in  the  exponent  of  the  unknown  quantity,  as  is  explained  by  an 
example  in  (Art.  163).  We  here  give  one  more  practical 
illustration. 

The  equation  x5  —  1=0,  is  a  binomial  equation  of  an  odd 
degree,  and  it  is  also  a  recurring  equation. 

It  is  also  divisible  by  (x  —  1),  having  apparently  but  one 
root,  -f-1.  The  division  produces 


THEORY  OF  EQUATIONS—  SUM  OF  COEFFICIENTS.    343 

O.  (1) 


a  recurring  equation  of  the  4th  degree. 
Dividing  again  by  x2,  we  obtain 

l=0.  (2) 


Assuming        fx+-\=P.     Then   z2+-L=P2—  2, 
\        x/  x 

And  (2)  becomes  P2+P — 1=0,         a  quadratic. 

Again:  Operating  under  (Art.  203),  we  obtain 


Add      * 


x 
Square  root, 

Or  x2  4-1(1^:^/5)2:=  —  1,        a  quadratic. 

Whence  the  five  roots  in  (x5  —  1)=0,  or  the  five-fifths  roots 
of  unity  are 

x=\,  or  x= 


or  s=i(,y5— 1— V— 

or  x= — 


or  a?=—  1(^5+1  +V—  10+2  V6)- 


EXAMPLES. 
1.  Find  the  roots  of  the  equation 


Ans.  Three  of  its  roots,  are  each  equal  1,  and  f.  _=h-  75  j 


2. 

3. 

See  (Art.  201).         Jws.  a?=±l,     or 
4.  a;4_|,r3_j_22;2--fa;+l=0. 

r=2,  or  £,  or  db^/  — 


344  ELEMENTS  OF  ALGEBRA. 

5.  5x<+8x*+9x*+8x+5=Q. 

(Art.  203)  reduces  this  to  the  following  quadratic  : 


6.  4*«—  24*5-f-57z4—  73z3-f57#2—  24*+4=0. 

Roots,      2,^l±^, 

7.  4#4  4-3*3  —  8z2  —3^+4=0. 

Ans.  #=rhl,  or  ~ 

o 

8. 


;=,  or  — 

Some  one,  or  more  of  the  foregoing  principles  will  apply  to 
the  solution  of  the  following  examples. 

9.  a?4—  2s3—  7aa—  824-16=0. 

Ans.  ar=l,  or  4,  or  —  ^f— 

10.  x  4  +2a:3  —  3^2  _4a:+4=0. 

^4ws.   ar=l,  1,  or  —2,  —2. 

11.  x*—  2x3—  25a:24-26a:+120=0. 

Ans.  #=3  or  5,  or  —  2,  or  —  4. 

12.  x* 


—  3. 
.  #=0,  or  1,  or 


13. 


.  .r=0,  or  4,  or  db^/  —  8. 

14.  o;3+5a:2-f3^—  9=0. 

^Iw5.  «=!,  or  —  3,  or  —  3. 

15.  a^Gz2—  7#—  60=0. 

^4tts.  ar=3,  —  4,  or  —  5. 

16.  23+8.r'-fl7;r-f-10=0. 

Ans.  x=  —  1,  or  —  2,  or  —  5. 

17.  x3—  29#3-|-198#—  360=0. 

Ans.  x=»3,  6,  or  20. 


THEORY  OF  EQUATIONS—  SUM  OF  COEFFICIENTS.     345 

—  27=0. 

Ans.  #=-£•,  -J,  or  27. 

—  37=0. 

Ans.  x=%,  -J,  or  37. 
20.  4x*-{-3x3  -\-ttx2  —  3.r-[-4=0. 


Ans.  x=—a±j+a2  and2a=(£drN/—  247). 
21.  *5—  13z4+67z3—  171*2+216z—  108=0. 

(See  Art.  198.)  Ans.  3,  3,  3,  or  2,  2. 


REMARK. — This  treatise  has  hitherto  omitted  one  process 
of  elimination,  which  Bourdon,  and  some  others,  have  intro- 
duced, and  which,  some  teachers  regard  as  very  scientific  and 
interesting.  We  cannot  quite  agree  with  this  class  of  teachers, 
for  whatever  is  most  simple,  and  most  practical,  we  regard  as 
most  scientific,  and  this  method  is  neither  very  simple,  nor 
practical.  Still,  in  a  work  like  this,  it  is  worthy  of  a  passing 
investigation  like  the  following. 

GENERAL    METHOD    OF   ELIMINATION   AMONG    EQUATIONS 
ABOVE    THE    FIRST   DEGREE. 

(Art.  205.)  Suppose  we  have  two  equations,  each  contain- 
ing x  and  y,  like  the  following : 

x+y-6=0.         (1))      Ans     x=4,  or  2. 
#3+2/3  —72=0.  (2)j  '    y=2,  or  4. 

To  make  the  illustration  clear,  we  give  the  values  of  #  and 
y.  Now  by  the  theory  of  equations  (Art.  156),  if  x=4,  equa- 
tion (1)  is  divisible  by  (# — 4),  without  a  remainder. 

That  is,  whatever  the  remainder  appears  to  be,  it  is  in  value 
zero. 

Dividing  (x-\-y — 6)  by  (x — 4)  the  quotient  is  1,  and  the 
apparent  remainder  is  (y — 2),  but  this  must  be  zero ;  therefore 
y=2,  corresponding  to  x=4. 

But  x=4,  corresponds  to  both  the  given  equations ;  there- 
fore equation  (2)  is  divisible  by  (x — 4),  and  (x — 4)  is- a 
common  measure  to  the  two  equations  (1)  and  (2). 


316  ELEMENTS  OF  ALGEBRA. 

Now,  therefore,  if  we  take  equations  (1)  and  (2),  and  ope- 
rate with  them  for  common  measure,  as  taught  in  (Art.  27), 
the  last  remainder  must  be  zero  ;  and  in  fact  we  shall  show, 
that  each  and  every  remainder,  throughout  the  whole  operation,  must 
be  zero.  And  if  we  take  care  that  only  one  letter  shall  appear 
in  the  last  remainder  —  that  is,  operate  until  one  letter  only 
appears  in  the  remainder  —  that  remainder  can  be  put  equal  to 
zero,  and  the  desired  elimination  is  effected. 

OPERATION". 

(A) 

y  3  —  72      (B)  remains. 

Observe  that  the  divisor  is  in  value  zero,  and  it  being  mul- 
tiplied by  x2  ,  the  product  (A)  must  be  zero  in  value  ;  and  that 
product,  subtracted  from  the  dividend  (zero),  the  remainder 
(B)  must  be  zero. 

Thus  we  might  establish  the  fact,  that  each  remainder  (and 
in  fact  each  product  also),  throughout  the  operation,  must  be 
zero. 

Continuing  the  operation  we  obtain 


—  x*y  —  xy2  -\-6x2  +6y*  —  72.         2d  remainder. 
Or,       —  xy(x-\-y)-\-Q(x2-\-y2)  —  72     =2d  remainder. 
But  we   learn   by  equation   (1)   that  (x-{-y)=6,   therefore 
we  can  divide  this  second  remainder  by  6,  and  we  have 


Qx 


6x — 2xy-\-y2 — 12  3d  remainder. 

Or,       (6 — 2y)z-f-y2 — 12  3d  remainder. 

(6 — 2y  )#-]-(  6 — 2y)(y — 6) 


y2— 12— (6— 2y)(y— 6).  4th  remainder. 
This  last  remainder  does  not  contain  x,  therefore  that  letter 
is  completely  eliminated,  and  the  remainder  put  equal  to  zero 
will  give  the  values  of  y.- 


THEORY  OF  EQUATIONS—  SUM  OF  COEFFICIENTS.     347 

N.  B.  —  When  this  method  is  clearly  analyzed,  it  will  be 
found  to  be  nothing  more,  than  elimination  by  addition  and 
subtraction. 

Here  we  have  two  equations,  (1)  and  (2),  each  equal  to 
zero,  and  we  multiply  one  of  them,  and  subtract  the  product 
from  the  other.  The  remainder  is  an  equation  equal  to  zero, 
and  if  its  terms  have  a  common  divisor,  we  can  divide  the 
equation  by  it.  Hence  we  have  no  new  principle,  nothing  to 
gain  by  keeping  these  forms  in  view.  The  common  method 
gives  us  much  more  freedom  of  action. 

We  give  one  more  example. 

Given,  (f*+f  *)(*-y)-1220=0>  d      and 

And,  x2-\-y2-^-x  —  y    —  132=0[ 

Assume,        %2-}-y2=P,     and  x  —  y=Q» 
Then,  P+Q—  132=0, 

And,  PQ—  1220=0  (Q, 


—  02-f  1320—  1220=0,  a  quadratic. 

We  extract  the  following  from  Prof.  Perkin's  Algebra. 
Indeed,  we  have  observed  it  in  several  other  works,  and  it  is 
given  as  an  appropriate  example  to  illustrate  this  method  of 
elimination. 

We  give  it  to  show  the  utter  inutility  of  this  mode  of  ope- 
ration, in  a  practical  point  of  view. 

Given,  a.*+Xy+y*—l=0,  (1) 

And,  .^+y3=0,  (2) 

to  find  a  single  equation  in  terms  of  y. 

FIRST  OPERATION. 
1)  x*+y*(x—  y 


—  x2y—  y2x  -f-  y—y 


—  y=first  remainder. 


348  ELEMENTS  OF  ALGEBRA. 

SECOND  OPERATION. 

/2—  '  !(*—  (2y3—  2y) 


_(2t/ 


second  remainder. 

Whence,         4y6 — 6y4-|-3y2 — 1=0,  is  the  equation  sought. 
As  the  sum  of  these  coefficients  equal  zero,  therefore,  y=l, 
and  if  it  were  not  for  this  lucky  circumstance  the  values  of  x 
and  y,  in  (1)  and  (2)  would  still  be  far  away. 

We  hare  observed,  that  all  remainders  in  these  operations, 
must  be  zero  ;  therefore,  our  first  remainder  must  be  zero,  and 
x—y— 2y3. 

Or,  x=(l — 2y2)y.  This  value  of  x  put  in  (2)  and  we 
have  (l-f2y2)32/3-f?/3=0.  Or,  (1— 22/2)3+l=0. 

Expanding  and  dividing  by  2,  produces 

4y6_62/4_|_3y2_1_o.     The  same  as  before. 
But  if  the  object  is  to  solve  equations  (1)  and  (2),  and  find 
the  values  of  x  and  y,  the  common  method  is  incomparably 
the  best. 

sa-H*y+ya  =  i.  (1) 

Equation  (2)  is  divisible  by  (#-f-y),  and  the  other  factor  is 
x2 — xy-\-y2 .         Therefore  we  have 

x+y=o.  (2) 

Or,  xt—xy+yV^Q,  (3) 

Subtracting  (3)  from  (1)  and  we  dbtain 
2ary=l,  or,  xy=^. 

This  last  added  to  (1),  and  subtracted  from  (3)  will  produce 

And, 


A.lso,  x= — 1,  andy=l. 


APPENDIX. 


DESIGNED     TO     SHOTf     THE     UTILITY    OF    ALGEBRA    IN     PHILOSOPHICAL 
INVESTIGATIONS. 


INEQUALITY. 

Expressions  of  inequality  sometimes  occur  in  the  mathematics,  but  we 
can  always  make  equations  of  them  by  adding  a  symbol  to  the  smaller 
quantity.  Then  we  can  reduce  the  expressions  the  same  as  equations. 
without  danger  of  confusion. 

EXAMPLES. 
1.  Find  the  limit  of  the  value  of  x  in  the  inequation 


Let  h  represent  the  difference  between  these  two  expressions, 

en, 

Whence, 


then,    .....      lx—  8?=?f+5+A,  is  a  perfect  equation 
o       o 


or 
That  is,  x  is  greater  than  the  number  2. 

2.  The  double  of  a  number  diminished  by  5  is  greater  than  25,  and  triple  of 
the  number  diminished  by  1  is  less  than  the  double  increased  by  13.     What 
numbers  will  satisfy  the  conditions? 

Let  ar=  the  number,  and  h  and  k  the  difference  of  the  expressions. 

Then,    ......          2;c—  5=25-^.  (1) 

3x—  7-t-&—  2*4-13.  (2) 

From     ....     (1)    *=15-H/i.     From     (2)    *=20—  k. 

That  is,  x  must  be  more  than  15  and  less  than  20,  consequently,  16,  17, 
18  and  19  will  each  answer  the  conditions. 

3.  The  sum  of  two  numbers  is  32,  and  if  the  greater  be  divided  by  the  less, 
the  quotient  will  be  less  than  5,  but  greater  than  2.     What  are  the  numbers? 

Let  x=  the  greater,  y=  the  less,  then  ar-f-y=z32.  (1) 

And    ....    f=5-nft.      (2)  f=3+Jfc.  (3) 

y  y 

From  (2)     x=5y—hy.     This  put  in  (1)  gives  6y—hy=32. 

on 

Therefore,    y=_.     From  (3)  x=2y+ky;  this,  put  in  (1), 


32 
gives  JF=£f£ 

From  these  last  equations  we  perceive  that  32  divided  by  less  than  6  will 
give  the  limit  of  y  in  one  direction,  and  divided  by  more  than  3  will  give 
its  limit  in  the  opposite  direction;  that  is,  y  must  be  more  than  5  and  less 
than  10  ;  consequently  x  must  be  more  than  22  and  less  than  27. 

349 


350  -  APPENDIX. 

'  DIFFERENTIAL  METHOD  OF  SERIES. 

A  series  purely  of  the  arithmetical  order  has  been  fully  treated,  as  also 
those  of  the  geometrical  order;  but  a  series  in  general  exist,  when  each  of 
the  terms  is  derived  from  one  or  more  of  the  preceding  terms,  according  to 
some  definite  law.  (Art.  141  ) 

Therefore,  a  variety  of  series  may  exist,  neither  arithmetical  nor  geo- 
metrical, and  these  can  be  investigated  by  means  of  finite  differences;  and 
the  word  differential  at  the  head  of  these  remarks,  is  not  identical  with  the 
same  word  applied  to  the  differential  calculus.  For  example, 

1,    5,     15,    35,     70,     126, 

and  so  on,  is  a  regular  series,  and  if  we  wished  to  find  its  10th,  12th,  or 
nth  terms,  it  would  be  too  unscientific  and  tedious  to  lind  it  through  the 
succession  of  terms  ;  we  must,  therefore,  investigate  for  general  principles. 

The  series  is          1,     5,     15,     35,     70,     126,     210, 

1st  order  of  diff.      4,     10,    20,    35,     56,      84, 

2d      »      «     «  6,    10,    15,    21,      28, 

3d      "      "     "  4,      5,       6,      7, 

4th     "      "     "  1,      1,     1, 

5th    "      "     «  0,    0, 

Here  it  will  be  observed  that  the  first  term  of  the  series  is  1.  The  first 
term  of  the  first  order  of  differences  is  4.  The  first  term  of  the  2d  order 
of  differences  is  6.  The  first  term  of  the  3d  order  of  differences  is  4.  The 
first  term  of  the  4th  order  is  1  j  and  lastly,  the  first  term  and  all  the  terms 
of  the  5th  order  are  0. 

To  obtain  general  principles,  however,  we  must  take  general  notation 
as  follows  :  (The  symbol  a4  is  read  a  sub.  4,  and  so  on  for  other  like 
symbols.) 

Let  .....     a,,      a2,      a3,      <*4,      a^     aa,     be  a  series. 

ij,      62,      63,      &4,      6a,          1st  differences 
ci»      C2»      cs'      C4'  2<*  differences. 

dlt     d2,     cZ3,  3d  differences. 

&c.  &c. 

By  this  notation  we  perceive  that  d^  would  represent  the  4th  term  of 
the  third  order  of  differences,  d  being  the  third  letter  of  the  alphabet. 

From  the  foregoing  notation,  we  readily  derive  the  following  equations  : 


and  so  on,  and  so  on,  and  so  on. 

By  transposition  (writing  the  letters  in  alphabetical  order),  we  have 


and  so  on,  and  so  on,  and  so  on* 

In  the  2d  equation  of  the  first  column  we  will  substitute   the  values  of 
an  I  6^,  then  we  shall  have  fl3=ff14~'W>i~hcit  (1) 


DIFFERENTIAL  SERIES.  351 

By  the  same  process  we  find 

^i-H'i-M-  (2) 

In  the  3d  equation  of  the  first  column  we  find 


Substituting  the  values  of  a3,  63  from  equations  (1)  and  (2)  gives  us 

a4=a1-f361-|-3c1-t-rf1.  (3) 

By  the  same  process, 

64=6,-i-3crt-3<VK  (4) 

Also  by  referring  back  we  find 

a5=a4-H4- 
Substituting  the  values  of  a4  and  b4  from  (3)  and  (4),  we  find 

(5) 


By  inspecting  equations  (1),  (3)  and  (5),  we  perceive  that  the  coefficients 
in  theSd  members  correspond  to  binomial  coefficients,  and  if  we  should  con- 
tinue the  process  to  any  degree,  they  would  still  correspond,  and  the  letters 
represent  the  first  terms  of  the  several  orders  of  differences. 

Therefore,  in  general, 


And      .     .     6n+l=H-nc-f-n  .  ^~d-\-  n  .  "f  .  ?=-V}-    &c. 

We  drop  the  sub.  ones  in  the  second  members. 

The  series  terminates  with  that  order  of  difference  which  becomes  con 
etant  ;  for  the  succeeding  order  becomes  zero. 

EXAMPLES. 

1.  Required  the  12th  term  of  the  series. 

1,     5,      15,    35,    70,     126, 
4,     10,    20,    35,    56, 
6,      10,     15,     21, 
4,     5,      6.. 

1,     1. 
Here        a=l,    6=4,    c==6,    rf=4,    «=1,    /=0,    n=ll. 

Whence,       a12=l-{-ll  .  4+11  .  £  .  6+11  .  {?  .  L  .  4-J-ll  .  5    |.f 
Or,    ...    al2=l-f-44-j-330-f-660-t-330=1365,    Ans 

2.  Required  the  15th  term  of  the  series 

1,  4,     10,    20,     35,    &c.  Ans.  670 
3    Required  the  20th  term  of  the  series 

6,    10,    15,    21,    Ac.  Ans.  253. 

4.  Required  the  nth  term  of  the  series 

1,    3,    6,    10,    Ac.  AntJ 

5.  Required  the  nth  term  of  the  series 

2,  6,     12,    20,     30,     <fe«.  Ans.  n"-\-n. 
N.  B.     To  solve  the  4th  and  5th  examples,  write  (n  —  1)  for  n  in  the 

formula. 


352  APPENDIX 

6.  Find  the  20th  term  of  the  series  1,  8,  27,  64,  125,  Ac.      Ans.  8000. 

7.  Required  the  50th  term  of  the  series 

1,     3,     6,     10,     15,    Ac.  Ans.  1275. 

Compare  examples  4  and  7. 

Our  investigation  thus  far  has  been  limited  to  finding  a  particular  term 
of  a  series. 

We  now  propose  to  find  the  sum  of  any  proposed  number  of  consecutive 
terms. 

Let  .     .     .     Oj,     a^,     a3,     a4,     a.,  .  .  .  an,     <fec.,  represent  a  series. 

With  the  terms  of  this  series  form  a  new  series  commencing  with  0, 
for  its  first  term  ;  adding  the  succeeding  term  of  the  proposed  series  to  the 
preceding  term  of  the  new  series,  thus, 

0,     (0+flj),     (0+arf-a2),         (0+flrf-a2-{-a3),     and  so  on. 

Here  it  is  obvious  that  the  (n-{-l)th  term  of  this  new  series  is  the  sum 
of  n  terms  of  the  proposed  series.  Therefore,  we  will  find  the  (/i-|-l/lh 
term  of  the  new  series  by  the  formula,  and  that  will  be  n  terms  of  the 
proposed  series  as  required. 

EXAMPLES. 

1.  Find  the  sum  of  10  terms  of  the  series  3,    5,    7,    9,     11,     Ac. 
The  new  series  to  be  formed  is 

0,     (O-j-3),     (0-f3-l-5),     (0-f3-}-5-|-7),     (0-j-3-{-5-f  7+9),     Ac 
That  is,     ....     0,     3,     8,     15,    24,     Ac. 

We  must  now  find  the  llth  term  of  this  series,  which  will  be  the  sum 
of  10  terms  of  the  proposed  series  as  required.  Ans.  120. 

2.  Find  the  sum  of  20  terms  :  also  the  sum  of  n  terms  of  the  series 

1,     3,     6,    10,    15,     Ac. 

Ans.  20  terms  =1540,  n 

3.  Find  n  terms  of  the  series  2,     6,     12,    20,    30,    Ac. 

Ans. 

4.  Find  the  sum  of  n  terms  of  the  series  of  cubes 

P,     23,     33,     43,     &c.  Ans. 

5.  Find  the  sum  of  10  terms,  also  the  sum  of  n  terms  of  the  series 

1,    4,    9,    16,    25,    36,    Ac. 
N.  B.     The  new  series  is     0,     1,    5,     14,    30,    55,     91,     &c 

Ans,  Sum  of  10  terms  =385.     Sum  of  n  terms  =L2re2+3ra+1)ra. 

6 

The  series  of  numbers  in  example  2  are  called  triangular  numbers,  be- 
cause they  may  be  represented  by  points  forming  equilateral  triangles,  thus: 


•       •  •••  t       •       •       * 

36  10 


DIFF1  RENTIAL  SERIES.  353 

In  like  ma.iner  the  square  numbers  in  example  5  can  be  represented  by 
points,  thus: 


In  forts  and  armories  shot  and  shells  are  piled  on  triangular,  square,  or 

oblong  bases. 

When  the  base  is  a  triangle,  and  the  pile  complete,  the  number  of  balls 

in  the  pile  is  expressed  by  the  answer  to  example  2,  in  which  re  represents 

the  number  of  layers  in  the  pile. 

When  the  base  is  a  square,  and  the  pile  complete,  the  number  of  shot  in 

the  pile  is  expressed  by  the  answer  to  example  5,  in  which  n  represents  the 

number  of  layers  in  the  pile. 

When  the  pile  is  an  oblong  and  complete,  the  upper  layer  is  a  single  row. 

The  next  larger  has  two  rows  and  one  more  ball  in  each  row.     The  third 

layer  consists  of  three  rows,  and  each  row  one  more  ball  than  the  preceding, 
Now  if  the  top  layer  consists  of    .....    m-j-1     balls, 
The  2d  must  consist  of  ...  2(m+2)     or    2ro+4      " 
The  3d   must  consist  of   ...  3(m+3)     or    3m+9       " 
The  4th  must  consist  of  .     .     .  4(m+4)     or    4m-4-l6     " 
The  5th  must  consist  of  ...  5(m+5)     or     5m+25     "    <fcc. 

The  number  of  balls  in  the  whole  pile  must  then  consist  of  the  sum  of 
the  series  (m+1),  (2w+4),  (3m-J-9),  <fec.,  to  n  terms  n  representing  the 
number  of  layers  in  any  pile. 

We  therefore  require  the  sum  of  this  series  to  n  terms. 

The  new  series  to  be  formed  is 

0,     (0+m+l),     (0+3771+5),     (0+6/71+14),    <fcc. 

1st  diff.        (m+1),     (2771+4),     (3m+9),     (4771+16),     &c. 

2d    diff.  (>H-3)>     (»»+5),     (wi+7). 

3d    diff.  2  2 

4th  diff.  0 

Now  by  the  formula,  the  (n+l)th  term  of  this  series  is 

an+1=0+n(m+l)+n  .  ^.(w+^+n  .  n~l  .  n-^X2. 

The  right  hand  member  of  this  equation  is  the  general  formula  for  an 
oblong  p'le  of  shot  or  shells,  in  which  (wi+1)  represents  the  number  in  the 
top  row,  and  n  the  number  of  layers. 


We  have  previously  found  that  rciw+I)  (M+^)  represents  a  triangular 

pyramidical  pile  and,    n(  n  "f"  w+_)  represents  a  square  pyramidical  pile 
6 

EXAMPLES. 

1  .  How  many  shot  are  in  a  triangular  pyramidical  pile  consisting  of  1  8 
layers?  Ans.  1140. 

2.  How  many  shot  or  shells  in  a  square  pile  consisting  of  15  layers  ? 

Ans.  1240. 

3.  The  top  row  of  an  oblong  pile  of  shot  or  shells  consists  of  31  balls,  and 
the  number  of  layers  is  30.     How  many  balls  are  there  in  the  whole  pile  ? 

30  An 


354  APPENDIX. 

4.  There  is  an  oblong  pile  of  balls  consisting  of  20  layers,  and   644* 
balls  ;  how  many  balls  are  in  its  base  ?  Ans.  740. 

5.  A  square  pile  of  shells  consists  of  12  layers,  the  upper  layer  has  8 
shells  on  a  side  ;  how  many  shells  are  in  the  pile  ?  Ans.  2330. 

6.  The  number  of  balls  in  a  triangular  pile  is  to  the  number  in  a  square 
pile  (having  the  same  number  of  layers)  as  5  to  9  j  required  the  number 
in  each  pile.  Ans.  455  and  819. 

Our  object  is  now  to  express  the  several  orders  of  differences  by  the  terms 
of  the  series, 

ai  a,  a3  «4  fl5  °6  &C' 

1st  order,  (a—  aj,    (a3—  «2),    (a—  «3),    (a,—  a4)  &c. 

2d  order,  (a3—  2^+^),    (a—  2«3+a2),    (a—  2a4+a3). 

3d  order,  («4—  3a3+3a2—  a,),    (a^—^a.^a—a  ) 

4th  order,  (a.  —  4a^-r-6a3  —  ia-j-Cj). 

Here  it  can  be  observed  that  the  coefficients  correspond  to  those  of  the  powers 
of  a  binomial. 

Also  observe  that  to  compose  thejirst  term  of  the  2d  order  of  differences, 
we  must  use  the  first  three  terms  of  the  series.  To  compose  the  first  term 
of  the  3d  order,  we  must  use  the  first  four  terms  of  the  series.  To  com- 
pose the  first  term  of  the  4th  order  of  differences,  we  must  use  the  first 
five  terms  of  the  series  :  and  in  general,  to  compose  the  first  term  of  the 
nth  order  of  differences,  we  must  use  the  first  (n-f-1)  terms  of  the  series. 
Observing  these  facts,  the  first  term  of  the  5th  order  of  differences  of  the 
preceding  general  series,  must  be  expressed  thus  : 

o6—  5a.+lOa4—  10a3+5«2—  aj. 

If  the  first  term  of  any  order  vanishes,  or  becomes  very  small,  the 
expression  for  it  may  be  put  equal  to  zero,  and  any  term  of  the  series  com- 
prised in  it,  can  be  found,  provided  the  other  terms  are  given. 

For  example,  suppose  the  4th  order  of  differences  in  a  series  becomes  0, 
then 


Now  suppose   the  3d  term   of  the  series  lost  or  unknown,  the  others 
being  given,  it  is  found  thus  : 


__ 

The  series  1,  8,  a,  G4,  125,  has  lost  its  third  term,  its  4th  order  of  dif- 


ference vanishes  ;  what  is  that  lost  term  ? 

Ans. 


6  6 

One  of  the  most  important  applications  of  this  calculus  of  finite  differ- 
ences is,  that  of  inserting  between  the  terms  of  a  series  some  new  term 
or  terms  subject  to  the  same  law.  This  is  called 

INTERPOLATION. 

We  have  already  seen  that  the  general  equation  is 

n  —  1.  ,  _     n  —  1     n  —  2 

T 
This  series  will  apply  to  any  term  beyond  a.     If  we   required  the  3d 


INTERPOLATION. 


355 


term  beyond  a,  we  must  put  n=3.     If  the  4th  beyond  a,  we  put  n=4, 
and  so  on. 

If  part  of  a  term  beyond  a,  we  put  w=  to  that  fractional  part. 

EXAMPLE . 

Given  the  logarithms  of  102,  103,  104,  and  105,  to  find  the  logarithm 
of  103.6,  which  is  the  1.6th  term  beyond  a. 


NO. 

LOGARI'MS. 

1ST  DIFF. 

2o  DIFF. 

3D  DIFF. 

102 
103 
104 
105 

2.  0086002 
2.  0128372 
2.  0170333 
2.  0211893 

.0042370 
.0041961 
.0041560 

—.0000409 
—.0000401 

—.0000008 

a=2.0086002,     6=  .0042370,     c—  —  .0000409,     d—  .0000008,      n=1.6 

a 2.  0086002 

nb +0.  0067792 


— 1 


c —  0.  0000196 


Ld  .......  +0.  00000005 


log.  103.6  2   01535985 

In  most  cases  the  3d  difference  may  be  omitted  :  in  this  case  it  only  in- 
fluenced the  8th  decimal  place.  If  we  were  to  require  the  logarithm  of 
102.3,  then  n=0.3.  If  104.7,  then  n=2.7,  and  so  on. 

Interpolation  is  much  used  in  astronomy,  as  the  following  example  will  show: 

At  noon  (Greenwich  time),  May  10th,  1846,  the  moon's  declination  was 
3°  27'  17.5"  north;  at  midnight  following  it  was  5°  46'  38"  north;  on 
the  llth,  at  noon,  7°  38'  57"  north,  and  the  following  midnight  10°  2' 
31.7"  north.  What  was  the  declination  on  the  10th  of  May  at  3h  P.  M? 

Here  a=3°  27'  17.5",  6=2°  19'  20.5",  c=—  T  1.5",  d——  1'  43.2",  and 
the  interval  of  12  hours  must  be  considered  as  the  unit  of  time  :  3  hours 
is,  therefore,  J  of  this  unit.  Hence,  n=0.25,  and  the  required  result  must 
be  the  sum  of  the  following  series  : 


The  1st  term  a  =3°  27'  17.5". 
2d  termn6  34'  50.1" 
3d  term  39.5". 

4th  term  —  5.6".' 

Sum~4°2'  41.5"T     Ans. 
To  show  tne  utility  and  application  of  this  formula,  we  give  the  following 

EXAMPLE  S. 

Given  the  logarithmic  sine  of  5°,  5°  12',  5°  24',  5°  36',  5°  48'  and  6°,  ft 
find  the  sine  of  any  or  all  intermediate  arcs. 

a. 

Sine  5°  0'=8.  9402360 
Sine  50  12'=8.  9572843 
Sine5°24'=8  9736280 
Sine5036'=8  9893737 
Sine5°48'=9  0045634 
Sine  GO  0'=9  0192346 


+6. 
1st  diff. 
.0169883 
.0163437 
.0157457 
.0151897 
.0146712 

2d  diff. 

6446 

5980 
5560 
5185 

-M. 

3d  diff. 

466 
420 
375 

4th  diff 

46 
45 

356 


APPENDIX. 


In  applying  the  formula  to  this  example,  12'  must  be  taken  as  the  unit 
of  space  between  the  terms.  Then  if  we  require  the  sine  of  5°  3',  we 
must  put  n=y3.2=2,  and  the  formula  gives 

Sine5°3'=8.9402960-r-2(.Ol69883)-H  .  tz? 


4G6=8.9402960+.0042471+604.3-f  25.6=8.9446061. 
Sine5°4'=8.9402960+i(.0169883)-H  .  tz? 

456=8.9402960+.0056628+716+29=8.9460333. 
For  the  sine  5°  13',  we  put  n=T12  and 
fin.  50  13'=8.9572843+T13(.0163437)+T12  .  j 

jti(420)=8.9572843+.0013619+S29+13=8.9586704. 

O 

To  find  the  log.  sine  of  5°  14'  13",  we  could  put  n=?I?=!??.     ' 

12      720 

The  work  of  finding  sines,  cosines,  and  tangents,  &c.,  is  already  done; 
and,  therefore,  in  respect  to  this  application,  the  formula  for  interpolation 
has  comparatively  lost  its  importance  ;  but  with  the  practical  astronomer, 
this  formula  is  of  the  greatest  importance,  and  must  forever  be  in  use. 

We  give  the  following 

EXAMPLE. 

For  the  English  Nautical  Almanac  the  moon's  declination  is  required 
for  every  hour  of  Greenwich  time;  but  to  compute  it  directly  from  the  lunar 
tables  and  trigonometry,  would  present  an  apalling  amount  of  labor,  and  to 
surmount  this  difficulty,  the  declination  is  computed  from  the  lunar  tables 
for  each  noon  and  midnight  of  Greenwich  time  ;  and  the  declination  in- 
serted for  the  intermediate  hours  by  interpolation. 

Opening  the  English  Nautical  Almanac  for  1854,  at  page  91,  I  find  the 
moon's  declination  as  follows  : 


3)  dec.  N.  +6 

May  23,  mean  moon,  6°  13'  14.4" 

May  23,  midnight,       8°  55'    6.4"  2°  41'  52.0" 

May  24,  noon,  11°  30'  26.2"  2°  35'  19.8" 

May  24,  midnight,     13°  57'  44.4"  2°  27'  16.2" 

May  25,  noon,  16°  15'  33.6"  2°  17'  49.2" 


6'  32.2" 
8'  3.6" 
9'  27.0" 


1'  31.4" 
1'  23.4' 


From  the  above  data  we  require  the  moon's  declination  at  the  commence- 
ment of  every  hour  between  noon  and  noon  of  May  23d  and  24th. 
For  the  first  hour,  or  1,  afternoon,  we  must  put  n=J  . 

The  D's  dec.  at  1  P.  M.  =6°  13'  14.4''+^  (2°  41'  52.0")+T2  .  itzl  X 


14.9"—  2.2"r=6°  26'  56.4"  North. 


SPECIFIC  GRAVITY. 


357 


The  following  are  the  results  taken  from  the  Nautical  Almanac  which 
serve  as  answers  to  twenty-two  examples  in  the  application  oi  this  formu-la 


Hours. 


f)Dec. 


Hours. 


Dec. 


1 

6C 

Ofi' 

565" 

13  

9° 

8'  19.3" 

2 

6 

40 

365 

14  

9 

21  29.4 

3 

6 

143 

15  

9 

34  36.6 

4 

7 

7 

498 

16 

9 

47  41.0 

5  

7 

91 

23.0 

17  

10 

0  42.5 

6  

7 

'M 

53.9 

18  

10 

13  40  9 

7  

7 

48 

22.3 

19  

10 

26  36.4 

8  .  .  . 

8 

1 

48.3 

20  

10 

39  28.7 

9  . 

8 

15 

11.7 

21  

10 

52  17.9 

10 

8 

32.6 

22  

11 

5  3.9 

11 

8 

-11 

50.8 

23  

11 

17  46.7 

12 

8 

rr 

64 

24  

..  11 

30  26.2 

SPECIFIC    GEAVITY. 

Gravity  is  weight.  Specific  gravity  is  the  specified  weight  of  one  body, 
compared  with  the  specified  weight  of  another  body  (of  the  same  bulk),  taken 
as  a  standard. 

Pure  water,  at  the  common  temperature  of  60°  Fahrenheit,  is  the  standard 
for  solids  and  liquids  ;  common  air  is  the  standard  for  gases. 

Water  will  buoy  up  its  own  weight.  If  a  body  is  lighter  than  water,  it 
will  float ;  if  heavier  than  water,  it  will  sink  in  water. 

If  a  body  weighs  16  pounds,  in  air,  and  when  suspended  in  water  weighs 
only  14  pounds,  it  is  clear  that  its  bulk  of  water  weighs  2  pounds  ;  and  the 
body  is  8  times  heavier  than  water ;  therefore  the  specific  gravity  of  this  body 
is  8,  water  being  1. 

If  the  specific  gravity  of  a  body  is  n,  it  means  that  it  is  n  times  heavier 
than  its  bulk  of  water.  Therefore  — 

If  we  divide  the  weight  of  any  body  by  its  specific  gravity,  the  quotient 
will  be,  the  weight  of  its  bulk  of  water. 

On  this  fact  alone  we  may  resolve  all  questions  pertaining  to  specific  gravity 

EXAMPLES. 

1.  Two  bodies,  whose  weights  were  A  and  B,  and  specific  gravities  a  and  b, 
were  put  together  in  such  proportions  as  to  make  the  specific  gravity  of  the 
compound  mass  c.  What  proportions  of  A  and  B  were  taken? 

A  quantity  of  water,  equal  in  bulk  to  A,  must  weigh  — 


A  quantity  " 


A+B 
c 


A  quantity  of  water,  equal  in  bulk  to  (A-^-B),  will  weigh 

A       B      A4-B 
Therefore,  -      )       = — ! — ;        Or,        bcA-\-acB=abA-\-abB ; 

Or,        b(o—  a)A=a(b— c}B. 


3-5?;  AITENDIX. 

•Hence  the  quantities  of  each  must  be  reciprocal  to  these  coefficients;  or  if  we 

a  /b  —  c\ 
take  one,  or  unity  of  B,  we  must  take  j  (  -  J   units  of  A. 

U     \,C"~~flyX 


2.  Hiero,  king  ef  Sicily,  sent  gold  to  his  jeweler  to  make  him  a  crown  ;  he 
afterwards  suspected  that  the  jeweler  had  retained  a  portion  of  the  gold,  and 
substituted  the  same  weight  of  silver,  and  he  employed  Archiwedes  to  ascer- 
tain the  fact,  who,  after  due  reflection,  hit  upon  the  expedient  of  specific  gravity 

He  found,  by  accurately  weighing  the  bodies  both  in  and  out  of  water,  that 
the  specific  gravity  of  gold  was  19,  of  silver  10.5,  and  of  the  crown  16.5. 
Fromthesedata  he  found  what  portion  of  the  king's  gold  was  purloined.  Re- 
peat the  process. 

The  preceding  problem  is  the  abstract  of  this,  in  which  A  may  represent  the 
weight  of  the  gold  in  the  crown,  B  the  weight  of  the  silver,  and  (./i-J-fi)  the 
weight  of  the  crown  ;  a=19,  6=10$,  c=16£. 

Then  if  we  take  J5=l,  one  pound,  one  ounce,  or  any  unity  of  weight,  of 

silver,  the  comparative  weight  of  the  gold  will  be  expressed  by  -j(  -  —  J. 

That  is,  for  every  ounce  of  silver  in  the  crown,  there  were  4  ^2j  ounces  of 
gold.  If  clearer  to  the  pupil,  he  may  resolve  this  problem  as  an  original  one, 
without  substituting  from  the  abstract  problem. 

3.  I  wish  to  obtain  the  specific  gravity  of  a  piece  of  wood  that  weighs  10 
pounds  ;  and  as  it  will  float  on  water,  I    attach    21  pounds  of  copper  to  it,  of  a 
specific  gravity  of  9.     The  whole  mass,  31  pounds,  when  weighed  in  water, 
weighs    only  4  pounds  ;  hence  27  pounds  of  the  31  were  buoyed  up  by  the 
water  ;  or  we  may  say,  the  same  bulk  of  water  weighed  27  pounds.     Require-1 
the  specific  gravity  of  the  wood. 

Let  s  represent  the  specific  gravity  of  the  wood. 

10 
Then  —  =  the  weight  of  the  same  bulk  of  water. 

21 

And  —  =   the  weight  of  water  of  the  same  bulk  as  the  copper. 
y 

10     7  30 

Hence,  ........    —-{--=27,         Or,        s==^=.405,  nearly  , 

4.  Granite  rock  has  a  specific  gravity  of  3.     A  piece  that  weighs  30  ounces, 
being  weighed  in  a  fluid,  was  found  to  weigh  only  21.5  ounces.     What  waa 
the  specific  gravity  of  that  fluid  ? 

The  weight  of  the  fluid,  of  the  same  bulk  as  the  piece  of  granite,  was  evi- 
dently 8.5  ounces.  Let  a  represent  its  specific  gravity. 

8.5  80 

Then  —  =  the  weight  of  the  same  bulk  of  water;  also  -Q=10=  tlio 

3  o 

weight  of  the  same  bulk  of  water. 

8.5 
Hence  —  =10,         Or,        s=.85   Ans.t  which  indicates  impure  alcohol 


MAXIMA  AND  MINIMA. 


5.  The  specific  gravity  of  pure  alcohol  is  .797 ;  a  quantity  is  offered  of  the 
specific  gravity  of  .85,  what  Proportion  of  water  does  it  contain? 

Let  A=  the  pure  alcohol,  and  W  =  the  water. 

Then 

Ans.  The  resolution  of  this  equation  shows  1  portion  of  water  to  2.255-j- 
portions  of  alcohol. 

6.  There  is  a  block  of  marble,  in  the  walls  of  Balbeck,  63  feet  long,  12  wide, 
and  12  high.     What  is  the  weight  of  it  in  tons,  the  specific  gravity  of  marble 
being  2.7  and  a  cubic  foot  of  water  62^  pounds,  Ans.     683  y\  tons. 

7.  The  specific  gravity  of  dry  oak  is  0.925 ;  what,  then,  is  the  weight  of  a 
dry  oak  log,  20  feet  in  length,  3  feet  broad,  and  2£  feet  deep  ?     Ans.  867  if  Ibs. 

We  may  now  change  the  subject,  to  make  a  little  examination  into  maxima 
and  minima.  For  this  purpose,  let  us  examine  Problem  2  (Art.  114). 

1.  Divide  20  into  two  such  parts  that  their  product  shall  be  140.     It  may 
be  impossible  to  fulfil  this  requisition,  therefore  we  will  change  it  as  follows : 

Divide  20  into  two  such  parts,  that  their  product  will  be  the  greatest  possible. 

Let  x-\-y—  one  part,     and    x — y=  the  other  part. 

Then  2.r=20,  and  #=10,  and  the  product,  x2 — y*,  is  evidently  the 
greatest  possible  when  y=Q.  Hence  the  two  parts  are  equal,  and  the  greatest 
product  is  100,  or  the  square  of  one  half  the  given  number. 

2.  Given  the  base,   m,  and  the  perpendicular,  n,  of  a  plane  triangle,  to  find 
the  greatest  possible  rectangle  that  can  be  inscribed  in  the  triangle.* 

Let  ABC  be   the   triangle,  BC=m,  A 

AF=n,  AD=x,  and  AE=x',  De  be  a 
very  small  distance,  so  that  x'  is  but  insen- 
sibly greater  than  x. 

As  Z),  comparatively,  is  not  far  from  the 
vertex,  it  is  visible,  that  the  rectangle 
a'b'c'd'  is  greater  than  the  rectangle  abed. 

If  we  conceive  the  upper  side  of  the 
rectangle  to  pass  through  D1,  in  place  of 
D,  and  we  represent  AD'  by  x,  and  At 
by  or',  it  is  visible  that  the  rectangle 
e'fg'h'  is  less  than  the  rectangle  efg  h. 

If  we  subtract  the  rectangle  abed  from  the  rectangle  a'b'c'd',  we  shall  have 
a  positive  remainder. 

If  we  subtract  the  rectangle  efg  h  from  the  rectangle  t'f'g'h',  we  shall  have 
a  negative  remainder. 

*  We  do  not  introduce  this  problem  to  show  its  solution  ;  it  bt.ongs  to  the  calculus, 
and,  in  its  place,  is  extremely  simple  ;  we  introduce  it  to  show  a  p-inciple  of  reasoning 
extensively  used  in  the  higher  mathematics,  and,  perchance  our  illustration  may  uj] 
a  pupil  in  his  progress  in  the  calculus. 


A  A  «• 


360  APPENDIX. 

The  rectangle  abed  cannot  be  the  greatest  possible,  so  long  as  we  can  have 
a  positive  remainder  by  subtracting  it  from  the  next  consecutive  rectangle 
immediately  below. 

After  we  pass  the  point  on  the  line  AF  where  the  greatest  possible  rectangle 
comes  in,  the  next  consecutive  rectangle  immediately  below,  will  become  less ; 
and  by  subtracting  the  upper  from  it,  the  difference  will  be  negative. 

Hence,  when  abed  becomes  the  greatest  possible  rectangle,  the  difference 
between  it  and  its  next  consecutive  rectangle  can  be  neither  plus  nor  minus 
but  must  be  zero. 

Therefore,  it  is  manifest,  that  if  we  obtain  two  algebraical  expressions  for  tho 
two  rectangles  abed  and  a'b'c'd',  and  put  their  difference  equal  to  0,  a  resolu 
lion  of  the  equation  will  point  out  the  position  and  magnitude  of  the  maximum 
rectangle  required. 

Put  the  line  ab=y,  and  a'b'=y'.  As  AD=x  and  AE=xl,  DF=n — x 
Wid  EF=n — xf.  The  rectangle  abcd=y(n — x),  and  a'b'dd'=y'(n — a:/). 

From  the  consideration  just  given,  the  maximum  must  give 

y'(n— -a;')—  y(n— #)=0. 

mx 


By  proportional  triangles,  we  have     x  :  y  : :  n  :   m        or.        y  =- 


n 


By  a  like  proportion,  we  have y=—x'. 

Put  these  values  of  y  and  yi  in  the  above  equation,  and,  dividing  by  — , 

we  have x'(n — x'}=x(n — x}  ; 

Or, x2—  x'*  =n(x—x'). 

By  division, x  -{-  x'   =n 

As  x'  is  but  insensibly  greater  than  x,  2x=n  ;  which  shows  that  AD  is  one 
half  AF,  and  the  greatest  rectangle  must  have  just  half  the  altitude  of  the  triangle. 

3.  Required  the  greatest  possible  cylinder  that  can  be  cut  from  a  right  cone. 

Conceive  the  triangle  (of  Prob.  2.)  to  show  the  vertical  plane  cut  through 
the  vertex  of  the  cone,  and  ab=y  the  diameter  of  the  required  cylinder.  Then, 
the  end  of  the  cylinder  is  .7854?/2,  and  its  solidity  is  .7854#2(n — *).  The  next 
consecutive  cylinder  is  .7854i/'2(?i — x").  Hence  y'2(n — a:')=;y2(n— ;r). 

By  similar  triangles  x  :  y  : :  n  :  m,       Or,        y  2= — ^—  and  y'2= — -z'9. 

Hence,       *'2(n— a:')=r2(n— *),         3r,        x3— *'3=n(z2— z'2) ; 
Divide  both  members  by  (z— z'),      and      »a-f  zz4-z'2==n(z-|-z'). 
As  x=x'  infinitely  near,  3z2=2wz,     or,    x=|n;  which  shows  tnat  the 
altitude  of  the  maximum  cylinder  is  5  the  altitude  of  the  cone. 

In  this  way  all  problems  pertaining  to  maxima  and  minima  can  be  resolved ;  bui 
the  notation  and  language  of  the  calculus,  in  all  its  bearings,  is  preferable  to  this.  We 
had  but  a  single  object  in  view — that  of  showing  the  power  of  algebra. 

(r  , 


/ 


XC  49563 


961661 


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